Chem 1045- Standard Enthalpies of Formation

If we calculate the change in enthalpy of a reaction by using heat of formation data, the correct calculation using the heats of formation of the reactants and products would be:

(sum of products) - (sum of reactants)

Consider the following chemical equations. C(s)+O2(g)→CO2(g)CO2(g)→CO(g)+12O2(g)ΔHΔH=−394kJmol=283kJmol Use the equations above and Hess's Law to calculate the standard enthalpy of formation for CO(g). The reaction for the formation of CO is represented by the equation below. C(s)+12O2(g)⟶CO(g)ΔH=?

-111 KJ/MOL

Given that the standard enthalpy of formations for NH3=−46.1kJmol, HCl=−92.3kJmol, and NH4Cl=−314.4kJmol, calculate the standard enthalpy change for the reaction. NH3(g)+HCl(g)⟶NH4Cl(s)

-176 KJ (sum of products- sum of reactants)

NH3(g)+HCl(g)⟶NH4Cl(s) Given the standard enthalpies of formation (below), calculate the enthalpy change for the reaction in kilojoules. NH3(g)HCl(g)NH4Cl(s)ΔH∘f=−46.1kJmolΔH∘f=−92.3kJmolΔH∘f=−314.4kJmol Your answer should have three significant figures.

-176 kJ

Given that CH4 (g)+Cl2 (g)→CH3Cl (g)+HCl (g) has an enthalpy change of −99.6 kJ, and CH3Cl (g)+Cl2 (g)→CH2Cl2 (g)+HCl (g) has an enthalpy change of −105.8 kJ, what is the change in enthalpy for CH4 (g)+2Cl2 (g)→CH2Cl2 (g)+2HCl (g)? Your answer should have one decimal place.

-205.4 KJ

Calculate ΔH∘298 in kilojoules for the process: Sb(s)+52Cl2(g)⟶SbCl5(g) Given the information below: Sb(s)+32Cl2(g)⟶SbCl3(g)ΔH∘298=-314 kJSbCl3(g)+Cl2(g)⟶SbCl5(g)ΔH∘298=-80.0 kJ Your answer should have three significant figures.

-394KJ

If the heat of formation for CO2 is −393.5kJmol, what is the change in enthalpy when 5.00 g of CO2 are formed? Your answer should have three significant figures. (Round your answer to first decimal place).

-44.7 KJ First, calculate the number of moles of CO2 using a formula mass of 44.009gmol from the periodic table: 5.00g×1.00mol44.009g=0.11361mol Next, calculate the total change in enthalpy: ΔH=0.11361mol×(−393.5kJmol)=−44.706kJ The answer should have three significant figures, so round to −44.7kJ.

In carbon combustion, the oxidation of solid carbon to produce carbon monoxide is represented below: 2C(s)+O2(g)⟶2CO(g) Use the following reaction enthalpies to find the enthalpy of the first reaction. C(s)+O2(g)2CO2(g)→CO2(g)→2CO(g)+O2(g)ΔHΔH=−394 kJ=+283 kJ Your answer should have three significant figures.

-505KJ 2×(C(s)+O2(g)⟶CO2(g))+(2CO2(g)⟶2CO(g)+O2(g))2C(s)+2O2(g)+2CO2(g)⟶2CO2(g)+2CO(g)+O2(g)2C(s)+O2(g)⟶+2CO(g) To find the reaction enthalpy, use the same operations on the given enthalpies. ΔH=2×(−394 kJ)+(+283 kJ)=−505 kJ Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.

How much heat is produced by the combustion of 125 g of acetylene (C2H2)? C2H2(g)+52O2(g)⟶2CO2(g)+H2O(l)ΔH∘=−1301.1 kJ/mol

-6.25* 10^3

Calculate the change in enthalpy associated with the combustion of 14.6 g of isooctane. C8H18(l)+25O2(g)⟶8CO2(g)+9H2O(l)ΔHc=−5461kJmol Report your answer with three significant figures.

-698 KJ

Use the reaction below to answer the question that follows. a.b.c.2Al(s)+32O2(g)⟶Al2O3(s)ΔH=−1676 kJ2Fe(s)+32O2(g)⟶Fe2O3(s)ΔH=−824 kJFe2O3(s)+2Al(s)⟶Al2O3(s)+2Fe(s) Calculate the change in enthalpy, in kilojoules for reaction "c" above. Your answer should have three significant figures.

-852 KJ

The standard heat of formation for HCl is −92.3kJmol. If 45.3 kJ of heat is released during the formation of HCl from its elements, how many grams of HCl have formed? Your answer should have three significant figures. (Round your answer to one decimal place).

17.9 g First, calculate the number of moles that must have formed using the information given. Note that a release of heat corresponds to a negative value of ΔH. −45.3kJ×1.00mol−92.3kJ=0.49079mol Next, calculate the number of grams of HCl, using a formula mass of 36.461gmol from the periodic table. 0.49079mol×36.461gmol=17.894g The answer should have three significant figures, so round to 17.9g.

In the heat of formation equation for CH3CH2OH, what would be the coefficient for hydrogen gas?

3

Calculate the ΔH∘298 for the equation: N2(g)+2O2(g)⟶2NO2(g) Use the equations below to help you calculate your answer: Step 1:N2(g)+O2(g)⟶2NO(g)ΔH∘298=180.5 kJ/mol Step 2:NO(g)+12O2(g)⟶NO2(g)ΔH∘298=−57.06kJ/mol

66.4 KJ/MOL

Complete the heat of formation equation for glucose (C6H12O6).

6C+ 6H2+ 302 TO C6H1206

Given that: 2NO (g)→N2 (g)+O2 (g)N2 (g)+O2 (g)+Cl2 (g)→2NOCl (g) has an enthalpy change of −180.6 kJ has an enthalpy change of 103.4 kJ What is the enthalpy change, in kilojoules, for 2NOCl (g)→2NO (g)+Cl2 (g)? Your answer should have three significant figures.

77.2 kJ

2C(s)+O2(g)2H2(g)+O2(g)H2O(l)→2CO(g)→2H2O(g)→H2O(g)ΔHΔHΔH=−222kJ=−484kJ=+44kJ Use the thermochemical data above to calculate the change in enthalpy for the reaction below. H2O(l)+C(s)→CO(g)+H2(g) Your answer should have three significant figures.

As shown below, we divide the first and second equations by 2. Notice that the second equation must be reversed, which will reverse the sign of the ΔH value of the second equation. Thus the equations are arranged as follows: 1/2(2C2(s)+O2(g)1/2(2H2O(g)H2O(l)C2(s)+1/2O2(g)+H2O(g)+H2O(l)H2O(l)+C(s)→2CO(g))→2H2(g)+O2(g))→H2O(g)→CO(g)+H2(g)+1/2O2(g)+H2O(g)→CO(g)+H2(g) Now we perform the same operations to the ΔH values for each equation to give the following: ΔH=1/2(−222 kJ)+1/2(484 kJ)+(44 kJ)=175 kJ Notice that coefficients in stoichiometric equations (indicating numbers of moles) are exact, so they do not constrain the number of significant figures.

"If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps." This is a statement of:

Hess's Law

What can be said about enthalpies of formation?

The sign depends on the reaction.

Hess's law can be used to accurately calculate the change in enthalpy of a reaction because:

enthalpy is a state function