Enthalpy

Calculate the number of kJ required for the reaction shown below, if you begin with 63 grams of calcium carbonate (CaCO3), and the ΔHrxn = +178 kJ. CaCO3(s) → CaO(s) + CO2(g)

+1.1 × 102 kJ

What is the enthalpy of the reaction N2O(g) + NO2(g) → 3 NO(g)? Substance -> Hf° (kJ/mol) N2O(g) 81.6 NO2(g) 33.2 NO(g) 91.3

159.1 kJ

What is ∆Hrxn for 2 Cr2O3(s) → 4 Cr(s) + 3 O2(g) given the enthalpy of formation of Cr2O3(s) is −1139.7 kJ/mol?

2279.4 kJ

What is Hrxn for Pb(NO3)2(s) → Pb(s) + N2(g) + 3 O2(g) given the enthalpy of formation of Pb(NO3)2(s) is −451.9 kJ/mol?

451.9 kJ

Calculate the number of kJ required for the reaction shown below, if you begin with 48 grams of oxygen (O2) and the ΔHrxn = +180 kJ. N2(g) + 2 O2(g) → 2 NO2(g)

+1.4 × 102 kJ

The enthalpy of formation of liquid ethanol (C2H5OH) is −277.6 kJ/mol. What is the equation that represents the formation of liquid ethanol?

2 C(s) + 3 H2(g) + ½ O2(g) → C2H5OH(l)

What is ∆Hrxn for 2 CHCl3(l) → 2 C(s) + H2(g) + 3 Cl2(g) given the enthalpy of formation of CHCl3(l) is −134.1 kJ/mol?

268.2 kJ

Calculate the number of kJ required for the reaction shown below, if you begin with 45.0 grams of water (H2O), and the ΔHrxn = +571.6 kJ. 2 H2O(l) → 2 H2(g) + O2(g)

714 kJ

The enthalpy of formation of gaseous carbon dioxide is −393.5 kJ/mol. What is the equation that represents the formation of gaseous carbon dioxide?

C(s) + O2(g) → CO2(g)

Which substance has an enthalpy of formation of zero?

Ca(s) or H2(g) or N2(g)

Identify the following reaction as an endothermic reaction, exothermic reaction, or both endothermic and exothermic: evaporation of water from your skin

Endothermic

Identify the following reaction as an endothermic reaction, exothermic reaction, or both endothermic and exothermic: PCl5(g) + 67 kJ → PCl3(g) + Cl2(g)

Endothermic reaction

Identify the following reaction as an endothermic reaction, exothermic reaction, or both endothermic and exothermic: dissolving NH4Cl in water to make a cold pack

Endothermic reaction

Identify the following reaction as an endothermic reaction, exothermic reaction, or both endothermic and exothermic: N2(g) + 3 H2(g) → 2 NH3(g) + 92.2 kJ

Exothermic reaction

Identify the following reaction as an endothermic reaction, exothermic reaction, or both endothermic and exothermic: burning a candle

Exothermic reaction

The enthalpy of formation of liquid water is −285.8 kJ/mol. What is the equation that represents the formation of liquid water?

H2(g) + ½ O2(g) → H2O(l)

Is this a true statement? The enthalpy of a reaction is the negative of the enthalpy of the reverse reaction.

Yes

Calculate the change in energy (in kJ) for the reaction shown below, if you begin with 12 grams of hydrogen (H2), and the ΔHrxn = −572 kJ. 2 H2(g) + O2(g) → 2 H2O(g)

−1.7 × 103 kJ

Calculate ∆Hrxn for 2 NOCl(g) → N2(g) + O2(g) + Cl2(g) given the following: ½ N2(g) + ½ O2(g) → NO(g) ∆Hrxn = 90.3 kJ NO(g) + ½ Cl2(g) → NOCl(g) ∆Hrxn = −38.6 kJ

−103.4 kJ

What is the enthalpy of the reaction 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g)? Substance Hf° (kJ/mol) H2S(g) −20.6 SO2(g) −296.8 H2O(l) −285.8 H2O(g) −241.8

−1036.0 kJ

Calculate ∆Hrxn for C2H4(g) + 6 F2(g) → 2 CF4(g) + 4 HF(g) given the following: H2(g) + F2(g) → 2HF(g) ∆Hrxn = 180.7 kJ C(s) + 2 F2(g) → CF4(g) ∆Hrxn = −680.0 kJ 2 C(s) + 2 H2(g) → C2H4(g) ∆Hrxn = 52.3 kJ

−1050.9 kJ

What is the enthalpy of the reaction 2 NO(g) + O2(g) → 2 NO2(g)? Substance -> Hf° (kJ/mol) NO(g) 91.3 NO2(g) 33.2

−116.2 kJ

Calculate ∆Hrxn for P4O6(s) + 2 O2(g) → P4O10(s) given the following: P4(s) + 3 O2(g) → P4O6(s) ∆Hrxn = −1640.1 kJ P4(s) + 5 O2(g) → P4O10(s) ∆Hrxn = −2940.1 kJ

−1300.0 kJ

What is the enthalpy of the reaction CH4(g) + 4 Cl2(g) → CCl4(l) + 4 HCl(g)? Substance -> Hf° (kJ/mol) CH4(g) −74.6 CCl4(l) −128.2 HCl(g) −92.3

−422.8 kJ

Calculate the change in energy (in kJ) for the reaction shown below, if you begin with 75.0 grams of oxygen (O2), and the ΔHrxn = −902.0 kJ. 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

−423 kJ

Calculate the change in energy (in kJ) for the reaction shown below, if you begin with 52 grams of fluorine (F2), and the ΔHrxn = −542 kJ. H2(g) + F2(g) → 2 HF(g)

−7.4 × 102 kJ

Calculate Hrxn for Ca(s) + ½ O2(g) + CO2(g) → CaCO3(s) given the following: Ca(s) + ½ O2(g) → CaO(s) ∆Hrxn = −635.1 kJ CaCO3(s) → CaO(s) + CO2(g) ∆Hrxn = 178.3 kJ

−813.4 kJ

What is the enthalpy of the reaction CH4(g) + 2 O2(g) → CO2(g) + 2H2O(l)? Substance -> Hf° (kJ/mol) CH4(g) −74.6 CO2(g) −393.58 H2O(l) −285.8 H2O(g) −241.8

−890.5 kJ