Chemistry II - Exam I (Ch. 11 & Ch. 12)

Question : At 25 degrees C, the solubility of NaCl in water is 359 kg/m^3. what is the solubility in g/mL? what is the solubility in mol/L?

(1 Kg = 1000g) 359 Kg = 359000 g 1m^3 = 1 x10^6mL (1 million) 359000g/1 x 10^6 mL = .359 g/mL --------------------------------------------------- 359 kG = 359000 g 1m^3 = 1 x 10^3 (1000) (mol of solute = mass/ molar mass of NaCl) 359000 g / 58.44 g/mol = 6143 mol (M = mol solute / L solution) 6143 mol / 1x 10^3 L = 6.14 mol/L

What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?

(14.5g/97.994 g/mol)/[(14.5g/97.994)+(125g/18.015)] = .1479/(.1479+6.938) = .1479/7.0859 = .0209 .0209 is the mole fraction of H3PO4 1-.0209 = .9791 .9791 is the mole fraction of water

Calculate ΔSdegree298 for the following change Zn(s) + CuSO4 (s) -> Cu(s) + ZnSO4 (S) Cu(s) = 33.15 ZnSO4 (s) = 110.5 CuSO4(s) = 109.2 Zn(s) = 41.6

(33.15+110.5)-(41.6+109.2) =-7.2 J/K

Additivity of ΔG Practice: In glycolysis, the reaction of glucose - GLY - to form glucose - 6 - phosphate - G6P -requires ATP to be present as described by the following equation: 1. Glu + ATP -> G6P + ADP ΔG298 = -17 KJ In this process, ATP becomes ADP, summarized by the following equation: 2. ATP -> ADP ΔG 298 = -30 KJ Determine the standard free energy change for the following reactions: Glu -> G6P ΔG298= ?

- Flip equation 2, the other way to say, times it by -1 so we can get equation 3. (-1)(equation2) = ADP -> ATP ΔGf = 30 KJ - Add equation 1 and 3. GLU + ATP -> G6P + ADP | -17 KJ + ADP ->ATP | 30 KJ ------------------------------ ----------- GLU -> G6P = 13KJ non spontaneous

Endothermic Dissolution

- Heat energy absorbed - Temp. decreases (think in inside)

The signs of Delta G indicates if a reaction will be spontaneous or not :

- If D. G<0, the reaction is spontaneous in the forward direction - if D.G > 0, the reaction is nonspontaneous in the forward direction - if D.G = 0, the system is at equilibrium

a solution contains 5.00 g of urea, CO(NH2)2, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 C is 23.7 torr, what is the vapor pressure of the solution (assuming ideal solution behaviors)

- Moles of Urea: 5g/60.06 g/mol = .0833 - Moles of water .100kg = 100 g 100/18.015 = 5.551 ---------------------- 5.551/(5.551+.0833)=.985 .985*23.7 mmHg = 23.3445 mmHg

Exothermic Dissolution

- energy released as heat - Temp. increases (think of exits)

Solutions

- homogeneous mixtures of two or more substances - homogenous and has uniform composition

Example: Assuming ideal solution behavior, what is the osmotic pressure of a solution of bovine insulin (molar mass 5700 g/mol) at 18 degrees C if 100.0 mL of the solution contains .103 g of insulin.

- mole = mass/molar mass = .103 g /5700 g/mol = 1.807 X10 -5 mol - M = mol solute / L solution = 1.807 x 10-5/.1000L = 1.807x10-4 mol/L -------------------------------------------------- i=1 R = .0821 atmL/molK T = 18 + 273 = 291 K M = 1.807x10-4 mol/L ----------------------------------------------------- II=iMRT = 1*1.807e-4*.0821*291 = 4.32 x 10-3 atm

Strong Acids examples

-HCL - HClO4 - strongest - HI - BHr - NHO3 - H2SO4

Example : Molality and Molarity What is the molality of 3.75 M aqueous ethanol - C2H5OH - solution ? The density of solution is .965 g/mL. Assume 1 Liter to start

-volume of solution : 1 L = 1000mL - mass of solution : d*v = .965 g/mL * 1000mL = 965 g - mole of solute : M*L = 3.75 M * 1 L = 3.75 mol - mass of solute : mole x molar mass ethanol = 3.75 mol * 46.07 g/mol = 172.8g -mass of water : mass of solution - mass of solute: 965 - 172.8 = 782.2 g -m = mol solute/Kg solvent = 3.75 mol/.7822Kg = 4.79 mol/Kg

Example: Without doing numerical calculations, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. 1. N2(g) + 3H2 (g) -> 2NH3(g) 2. HCl (g) + NH3 (g) -> NH4Cl (s) 3. (NH4)2Cr2O7 (s) -> Cr2O3(s) + 4H2O (g) + N2(g) 4. 2Fe (s) + 3O2(g) -> Fe2O3(s)

1. Low entropy 2. Low entropy 3. High entropy 4. low entropy

Factors that influence Entropy-Phase

1. phase of substance 2. The temperature of the substance 3. The type and number of particles that make up the substance 4. Variations in type of particles

Example: Additivity of ΔG Given: 1. P4(s) + 5O2(g) -> P4lO10(s) = -2697.0 kj/mol 2. 2H2(g) + 02 (g) -> 2H2O (g) = -457.18 kj/mol 3. 6H2O (g) + P4O10(s) -> 4H3PO4 (l) = -428.66 kj/mol Determine the standard free energy of formation, ΔGf, for phosphoric acid equation to represent ΔGf of phosphoric acid: 3/2 H2 + 1/4 P4 + 2O2 -> H3PO4

1.) 1/4 *3 => 3/2H2O+1/4P3O10->H3PO4 2.) 1/4*1=> 1/4P4+5/4O2->1/4P4O10 3.) 3/4*2=>3/2H2+3/4O2->3/2H2O ------------------------------------------ ΔGf 1.) 1/4 (-428.66) = -107.165 2.) 1/4 (-2697.0) = -674.25 3.) 3/4 (-457.18) = -342.885 ---------------------------------------------- Add all : 3/2 H2 + 1/4 P4 + 2O2 -> H3PO4 ΔGf = -1124.3 kj/mol

Entropy Change Practice Determine Delta S298 for the following: N(g) +O(g) -> NO (g) DeltaS298 = ?X N2(g) + O2(g) -> 2NO(g) DeltaS298 = 24.8 J/K N2(g) -> 2N(g) DeltaS298 = 115.0 J/K O2(g) -> 2O(g) DeltaS298 = 117.0 J/K

115.0+117.0+2X=24.8 2X=24.8-115.0-117.0 2X=-207.2 =-103.6

What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?

125 g = .125 kg 14.5 g H3PO4 * (1 mol/ 98 g H3PO4) = .148 mol H3PO4 .148 MOL / .125 Kg= 1.184

a KCL solution is 0.500 M. What is the molarity of ions in this solution 1 M 2M 4M

1M

immiscible

2 liquids that have low mutual solubility and do not mix - polar and non polar liquids - oil and water

At what temperature is the standard Gibbs free energy of formations reported?

25 Degrees C or -298.15 K

Spontaneous Exothermic Reaction example:

2H2 (g) + O2 (g) + SPARK -> 2H2O (g)

spontaneous example

2H2 (g) + O2 (g) -> 2H2O spontaneous

non spontaneous example

2H2O (l) -> 2H2 (g) + O2 (g)

Non spontaneous Definition

A process that DOES NOT occur naturally under a specific set of conditions

Spontaneous definition

A process that DOES occur naturally under a specific set of conditions

Entropy Practice: Predict signs of entropy: A. Pb2+ (aq) + S2- (aq) -> PbS(s) B. 2Fe (s) + 3/2 O2(g) -> Fe2O3(g) C. 2C6H14 (l) + 19 O2(g) -> 14 H2O (g) + 12CO2(g)

A. - because PbS is a pure solution and aq solutions have a higher entropy. Equation is low entropy B. - because gas disappeared which is a huge loss of entropy C. + because more gas particles in produce side, rise in entropy

Raoult's Law Calculations: The vapor pressure of methanol -CH3OH - is 94 torr at 20 degrees C. The vapor pressure of ethanol - C2H5OH- is 44 torr at the same temperature. A. Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol B. Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at 20 degrees C. C. Calculate the mole fraction of methanol and of ethanol in the vapor pressure above the solution

A. Xmethanol = (50.0g/32.05 g/mol) /((50.0g/ 32.05 g/mol) + (50g/46.08g/mol)) = 1.560/(1.560+1.085) = .590 Xethanol = 1 - Xmethanol = 1 - .590 = .410 B. Pmethanol = Xmethanol Pmethanol > .590 *94 torr = 55.4 torr Pethanol = XethanolPethanol> .410 * 44 torr = 18.0 torr C. Dalton's Law of Partial Pressure: Pa = XaPtotal -> Xa= Pa/Ptotal - XMethanol = Pmethanol/ (Pmethanol+Pethanol) = 55.4/(55.4+18.0) = .754 -Xethanol = 1 - Xmethanol = 1-.754 = .246

Which of the following would be a weak electrolyte in solutions? A. Sulfuric Acid B. Acetic Acid C. Hydroiodic Acid

Acetic Acid

Second Law of Thermodynamics states that

All spontaneous changes cause an increase in the entropy of the universe. -Delta Suniverse > 0 for a spontaneous process -DeltaSuniverse <0 for a nonspontaneous process -DeltaSuniverse =0 for a process at equilibrium

Example : Molarity, Molality, and Mole Fraction: Commerical Bleach is 5.25 % sodium hypochlorite by mass: 100.00 g of bleach contains 5.25 g of NaClO. Calculate the molarity (M), the Molality (m), and Mole Fraction (X NaClO) of commerical bleach. Predict the Van't Hoff Factor (i) for commercial bleach. The bleach solution has a density of 1.11 g/cm^3 , water has a density of 1.00g/cm^3

Assume there is one liter in solution: - volume of solution : 1 L = 1000 mL - Mass of solution : d * v = 1.11 g/ cm^3 * 1000mL (cm^3 = 1 mL) = 1110 g - Mass of solute : mass % * mass of solution = .0525 * 1110 = 5.8275 g - mass of water : 1110 - 5.8275 = 1104.1725 g = 1.104 kg -------------------------------------------------- Mol of Solute : mass/molarmass of NaClO = 5.8275 g / 74.44 g/mol = .07828 mol -------------------------- M = mol solute/ L solution = .07828 mol/1L = .0783 mol/L --------------------------------------------- m = Mol Solute/Kg solvent = .07828 mol/1.104Kg = .0709 mol/kG ---------------------------------- Xa = Mol A/Total moles = .07828/(.07828 +(1104/18.02 - mm of H20) = .00129 --------------------- Van't Hoff Factor of NaClO = 2

Which of the following would be a poor choice for making conductive solutions? -C6H12O6, CH3OH, CH3CH2OH -HCL,KBr, HNO3 -NaOh, KOH, LiOh

C6H12O6, CH3OH, CH3CH2OH

Henry's Law equation

Cg = k Pg Cg= solubility k= proportionally contant Pg = partial pressure

Entropy Example:

Dandelion dispersing their things

Direction of Spontaneity change A reactions have a delta H 298 = -50kj/mol and delta S 298 = -250 J/mol.K. Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous? If spontaneous, under what temperature conditions will it become spontaneous?

Delta G = Delta H - T DElta S Delta G= -50-298(-0.250)=24.5 kJ/mol [nonspontaneous] Delta G > 0 , non spontaneous: T= Delta H/ delta S= -50/(-0.25) = 200K

What is the delta S if the Delta H is +88.00 kJ and delta G is +11.21 kJ at 298 K ?

Delta G = DeltaH - Tdelta S - 11.21 kJ = 88.00 kJ - 298 (delta S) -Delta S = (88.00 - 11.21)/298 = .258 kJ = 258 J

remember Delta G sys = Delta H sys- T deltaS sys

Delta G sys = Delta H sys- T deltaS sys

Example : Vapor Pressure Lowering - Seawater Sea water is a solution of 3.0% (by mass) NaCl along with a small number of other salts. Calculate the DECREASE in vapor pressure of water at 25 degrees C caused by NaCl. The vapor pressure of pure water at 25 degrees C is 23.80 mmHg.

Delta P = Xsolute P^degree solvent - 3.0% (by mass): 3.0 g NaCl/100g solution 3.0 g NaCl/97.0g water Mole of 3.0 g NaCl = 3.0g/ 58.44g/mol = .0513 mol ---------------- note that one mole of NaCl produces 2 moles of solute particles. i = 2 -------------------- mole of ions = 2 * .0513 mole = .1026 mol mole of 97.0 g water = 97.0g/18.02g/mol = 5.383 mol ---------------------- Xion=moles of ions/total moles in sample : .1026/(.1026+5.383) = .0187 Delta P = XsolutePsolvent = .0187 * 23.80 mmHg = .45 mmHg

calculate Delta S 298 for the following change 2H2 (g) + O2 (g) -> 2H2O (l) --------------------------------- H2 (g) = 130.7 02 (g) = 205.2 H2O (l) = 70.0

Delta S rxn = (2*70.0)-(2*130.7)+(1*205.2) = -326.6 J/K

Arrange the following sets of systems in order of increasing entropy. Assume one mole and equal temperature: H2O (l), H2O (g), H2O (s)

H2O(s)<H2O(l)<H2O(g)

Spontaneous Endothermic Reaction example:

HC2H3O2 (aq) + NaHCO3 (aq) -> NaC2H3O2 (aq) + CO2 (g) + H2O (l)

Osmotic Pressure Equation

II=iMRT i: # of particles in solution M: molarity R: gas constant T: tempt in Kelvins

Units of entropies

J/mol K

Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene? Freezing point of benzene is 5.5 degrees C. Kf of benzene is 5.12 C/m.

Kf = 5.12 c/m i=1 m=? I2 molar mass = 253.82 -------------------------- (9.04/253.82)/.0755 = .4718 --------------------------- 5.12*1*.4718 = 2.4156 5.5 - 2.4156 = 3.1 degrees C

Strong Bases examples

LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)2, Sr(OH)2, Ba(OH)2 - all 1A metal hydroxides and heavy 2A metal hydroxides

m =

Mol solute / Kg solvent

Xa =

Moles of A / total moles in sample

Example : Vapor Pressure Lowering - Antifreeze Ethylene glycol - HOCH2CH2OH - the major ingredient in antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100 degrees C, the vapor pressure of pure water is 760mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass at 100 degrees C.

P = XwaterP^degree water - 30.2 % - > 30.2 g ethylene glycol / 100g sol - > 30.2 g ethylene glycol / 69.8 water g (100-30.2) - mole of 30.2g ethylene glycol = 30.2 g/ 62.07g/mol = .4865 mol - mole of 68.8 g water = 69.8g/18.02 g/mol = 3.873 mol - X water = moles of water/total moles in sample = 3.873/ (.4865+3.875) = .8888 P=XwaterPwater = .8888 * 760 = 675

Vapor-Pressure Lowering equation

P1 = X1P1^degree symbol

Osmotic Pressure (pi symbol) equation

Pi = iMRT

Assuming ideal solution behavior, what is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO3)2 in water at 25 degrees C? The volume of the solution is 275 mL. Molar mass of Ca(NO3)2 is 164.088 g/mol

Pi = iMRT i=3 T= 25+273 = 298 K R = .0821 275 mL = .275 L ------------------------------------ Pi*.275*164.088 = 3*1.64*.0821*298 pi*45.124=120.371 120.371/45.124 = 2.667

Boltzmann formula for entropy

S = KInW K = boltzmann constant (1.38 x 10-23 J/K) W= # of microstates

SHBrO4(g), SHBr(g), SH2(g) arrange the following sets of systems in order of increasing entropy, assume one mole and equal temperature

SH2(g)<SHBr(g)<SHBrO4(g)

Calculate ΔS degree 298 for the following change: SnCl4(l) - > SnCl4(g) Sdegree (J mol -1 K -1) SnCl4(g) = 366 SnCl(l) = 259

SnCl4(g) - SnCl(l) 366-259 =107 J/K

Solvent

Substance present in the LARGER amount (think of vent being big)

Delta G Example: To calculate the temperature at which the spontaneity of a reaction changes from... spontaneous to non spontaneous or non spontaneous to spontaneous, find the temperature at which delta G=0 what is the equation

T= Delta H/ Delta S

Which of the following is not a colligative property? boiling point elevation temperature of a solution freezing point depression

Temperature of a solution

Ionic Electrolytes

The electrostatic attraction between polar water molecules and ions from an ionic compound such as KCL ( are ION-DIPOLE INTERACTIONS)

Third Law of Thermodynamics

The entropy of a pure perfect crystalline substance at zero kalvin is zero

Van't Hoff Factor

The number of ions that each unit of electrolytes splits into, represented as i - i for NaCl is 2

Osmotic Pressure

The pressure at which solvent molecules move through the membrane at the same rate in both directions

Solute

The substance present in the SMALLER amount (think of something cute and small)

Mircrostate (W) Definition

a specific configuration of the locations and energies of the particles in a system

Third law of Thermodynamics define a reference point:

at 0 K, entropy of a perfect crystal is 0

boiling point elevation definition

at any given pressure, the boiling point is higher

Salt water freezes

below 0 degrees C

Entropy definition

dispersion of matter and energy

surroundings

external part of the system

V.H.F for MgCL2

i = 3

Example: Sucrose in water - What is the freezing point of a solution containing 115.0 g of sucrose (molar mass 342.3 g/mol) dissolved in 350.0 g water ? graph: Solvent: H20 Boiling Point: 100.0 degrees C Kb (c/m) : .512 Freezing Point : 0.0 C Kf (c/m): 1.86

i= 1 Kf= 1.86 c/m m = ? --------------------------------------- m= mol solute/Kg solvent = (mass/molar mass)/Kg solvent = (115.0g/342.3 g/mol)/.3500= .9599 mol/Kg ------------------------------ ΔTf=iKbm = 1*1.86*.9599 = 1.79C Tf= 0-1.79 = -1.79 Degrees C

Spontaneity definition

if a reaction is spontaneous in one direction, it will be non-spontaneous in the reverse direction under the same conditions

osmosis

is a diffusion-driven transfer of solvent molecules across a semi permeable membrane - osmosis causes the volume and level of the solution to rise

Pure water is a poor conductor because

it only slightly ionizes to yield hydronium ions and hydroxide ions. Within 555 million, only one ionize to ions

Thermal Population and Oxygen Solubility a certain species of freshwater trout requires a dissolved oxygen concentration of 7.5 mg/L. Could these fish thrive in a thermally polluted mountain stream (water temperature : 30.0 degrees C, Partial pressure = .17 atm). Henry's law constant from oxygen in water at 30.0 degrees C is 1.15x10-3 mol/L*atm

k = 1.15 x 10-3 mol/L*atm Pg = .17 atm ------------------- Cg = k Pg > 1.15 x 10-3 mol/L*atm *.17 atm = 1.96 x10-4 mol/L 1.96 x 10-4 mol/L * 32 g/mol (mm of oxygen) = 6.27 x 10-3 g/L convert to 6.27 mg/L It is not enough to keep trout alive. Too hot, O2 levels too low

A 12.0 g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at -1.94 degrees C. Assuming ideal solution behavior, calculate the molar mass of the substance. Kf of water = 1.86 c/m

kf= 1.86 i=1 m=? ΔTs = -1.94 ----------------------------------- 1.94/1.86*1 = 1.043 mol / kg 1.043 * 0.0800 kg = .08344 mol ->12.0 / .08344 = 143.8 or 144

M=

mol solute/L solution

solubility

of a compound in a particular solvent is the max concentration in solution that can be achieved. - solvent and temperature are important

unsaturated

solute < solubility

saturated

solute = solubility

supersaturated

solute > solubility increase the concentration of (a solution) beyond saturation point.

How do solutions differ from other mixtures?

solutions are homogeneous at the molecular level, while other mixtures are heterogeneous

miscible

some liquids mix with each other completely and have mutual solubility - both components usually = polarity (i.e. polar/polar or nonpolar/nonpolar) like dissolves like - water and ethanol

electrolytes

some substances dissolve in water and yield ions in solution

Formation of a solution is

spontaneous

Standard Entropy, DeltaS298

standard conditions - 1 atm and 1 M for solution at 298 K Entropies are always positive

System definition

subject of specific interest

nonelectrolytes

substances that do not yield ions and do not conduct an electric current in solution - most of molecular compounds (fe: H2O, glucose, C12H22O11)

vapor pressure lowering definition

the liquid-vapor curve beneath the solvent cure or curve

Freezing point depression definition

the solid-liquid curve for the solutions falls to the left to the pure solvent

HCL in water is

very good conductors

Isothermal

when no obvious temperature change from solute and solvent, before and after - for example : london-disperse force between He + Ar

What is the ΔS if the ΔH is +88.00 kJ and ΔG is +11.21 kJ

ΔG = ΔH -TΔS 11.21 = 88.00 -298 (ΔS) =(88.00-11.21)/298 = .257 kJ .257 kJ = 1000 = 257 J

use the standard free energy data to determine the free energy change for the following reactions, which is run under standard state conditions and 25 degrees C. Identify it as either spontaneous or non spontaneous at these conditions: Fe2O3(g) + 3CO (g) -> 2Fe (s) + 3CO2(g) Δ Gf (CO2, g) = -394.36 Kj/mol ΔGf (CO, g) = -137.15 Kj/mol Δ Gf (Fe2O3, s) = -742.2 Kj / mol

ΔG=∑vΔGf(products)−∑ΔGf(reactants) [2(0) +3(-394.36)]-[(-742.2)+3(-137.15)] =-29.43 Kj/mol Spontaneous

What is the difference between ΔS and ΔS^degree for a chemical change?

ΔS: any change in entropy under any set of conditions. ΔS degree: any change in entropy at standard conditions - 1 atm for gases, 1 M for solutions - if no temperature information is provided, assume the temperature is 298.15 K

entropy change of a reaction Delta S rxn can be obtained from S by: (equation)

ΔSrxn=∑nS(products)−∑nS(reactants) n= coefficent of the balanced equation when calculating Delta S

Boiling Point Elevation Equation

ΔTb=iKbm

Boiling-Point Elevation equation

ΔTb=iKbm

Example: What is the boiling Point of a solution containing 92.1 Got I2 and 800.0 g of chloroform - CHCl3. Assuming that I2 is non-volatile. Graph: Solvent: Chloroform Boiling Pt: 61.26 degrees C Kb (C/m) 3.63 Freezing pt: -63.5 C kf (c/m): 4.68

ΔTb=iKbm i= 1 Kb= 3.63 m = ? --------------------------- m= mol solute/Kg solvent > (mass/molarmass)/Kg solvent > (92.1/ 253.81)/.8000Kg > .4536 mol/Kg -------------------- ΔTb=iKbm 1*3.63*.4536 = 1.65 degrees C ΔTb= 61.26 (boiling pt) + 1.65 C= 62.91 degrees C

Freezing-Point Depression equation

ΔTf = iKfm

Practice: A 12.0g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at -1.94 degrees C. Assuming ideal solution behavior, calculate the molar mass of the substance. Kf of water = 1.86 c/m

ΔTf=iKbm > m = ΔTf / iKf -------------------------------- ΔTf = -1.94 i=1 Kf=1.86 c/m ---------------------------------- 1.94/(1*1.86)=1.043 mol/Kg - 1.043 mol/Kg * .0800 Kg = .08344 mol molar mass = mass / mol = 12.0 g/ .08344 mol = 144 g/mol

When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of ∆G, ∆H, and ∆S for this process

∆G negative, ∆H positive, and ∆S positive Explanation that I don't understand why but its here: ∆G is negative as the process is spontaneous. ∆H is positive as with the solution becoming cold, the dissolving must be endothermic. ∆S must be positive as the drives the process, and expected for the dissolution of nay soluble iconic compound.

A reaction has ∆Hº298 = -50 kj/mol and ∆Sº 298 = -250 J/mol.K. Is the reaction spontaneous at room temperature? if not, under what temperature conditions will it become spontaneous? if not spontaneous, under what temperature conditions will it become nonspontaneous.

∆Gº=∆Hº-T∆Sº (-50)-298(-.250) =24.5 kj/mol Non spontaneous ∆Gº > 0, T= ∆H/∆S -50/-.250=200 K B. The reaction is nonspontaneous at room temperature. It will become spontaneous when the temperature decreased until lower than 200 K

A reaction has ∆Hº 298 = 100 kj/mol and ∆Sº 298 = 250 J/mol.K. Is the reaction spontaneous at room temperature? If not, what temperature conditions will it become spontaneous?

∆Gº=∆Hº-T∆Sº 100 kj -298(.250) =25.5 nonspontaneous 100/.250 = 400 A. the reaction is not spontaneous at room temperature. It will become spontaneous when the temperature increased until higher than 400 K