Question
\(CaCO_3\) decomposes according to the balanced equation \(CaCO_3(s)→CaO(s)+CO_2(g)\).Based on the standard enthalpies of formation provided in the table above, what is the approximate enthalpy change of the reaction?
A −180kJ
B +180kJ
C −1460kJ
D +1460kJ
▶️Answer/Explanation
Ans:B
The enthalpy change for a chemical reaction can be estimated using \(\Delta H^{0}_{rxn}=\sum \Delta H^{0}_{f}{,products}-\sum \Delta H^{0}_{f}{ , reactants}\) Based on the reaction and the values provided,
\([ΔH_{rxn}]=[(1mole)\Delta H_{f}^0(CAO)+(1mole)\Delta H_{f}^0(CO_2)]-[(1mole)\Delta H_{f}^0(CaCO_3)]\)
\([-640kJ+(-390kJ)-(-1210kJ)]\rightarrow +180kJ\)
Question
Based on the information above, what is ΔH° for the reaction \(SO_2(g)+\frac{1}{2}O_2(g)→SO_3(g)\) ?
A−591.3kJ/molrxn
B −99.1kJ/molrxn
C +99.1kJ/molrxn
D +591.3kJ/molrxn
▶️Answer/Explanation
Ans:B
The standard enthalpy of reaction (ΔH°) is equal to \(\sum \Delta H^0_f,products-\sum \Delta H^0_f,reactants\) ; thus,ΔH°\(=(-359.2kJ/mol)-(-296.1 kJ/mol)=-99.1kJ/mol_{rxn}\).
Question
Natural gas consists primarily of \(CH_4\) , which is combusted according to the following chemical equation.
\(CH_4(g)+2O_2(g)→CO_2(g)+2H_2O(g)\)
Based on the standard enthalpies of formation in the table above, which of the following expressions give the approximate enthalpy change for the reaction \(\Delta H^0_{rxn\)?
A \(\Delta H^0_{rxn}=[(−75kJ/mol)−[(−390kJ/mol)+2(−240kJ/mol)]\)
B \(\Delta H^0_{rxn}=[(−75kJ/mol)−[(−390kJ/mol)+(−240kJ/mol)]\)
C \(\Delta H^0_{rxn}=[(−390kJ/mol)+2(−240kJ/mol)]−(−75kJ/mol)]\)
D \(\Delta H^0_{rxn}=[(−390kJ/mol)+(−240kJ/mol)]−(−75kJ/mol)]\)
▶️Answer/Explanation
Ans:C
The enthalpy change for a reaction can be estimated from \(\sum \Delta H^0_f,products-\sum \Delta H^0_f,reactants\) . Using the values given, \(\Delta H^0_{rxn}=[(−390kJ/mol)+2(−240kJ/mol)]−(−75kJ/mol)\).
Question
Natural gas consists primarily of \(CH_4\) , which is combusted according to the following chemical equation.
\(CH_4(g)+2O_2(g)→CO_2(g)+2H_2O(g)\)
Based on the standard enthalpies of formation in the table above, which of the following expressions give the approximate enthalpy change for the reaction \(\Delta H^0_{rxn\)?
A \(\Delta H^0_{rxn}=[(−75kJ/mol)−[(−390kJ/mol)+2(−240kJ/mol)]\)
B \(\Delta H^0_{rxn}=[(−75kJ/mol)−[(−390kJ/mol)+(−240kJ/mol)]\)
C \(\Delta H^0_{rxn}=[(−390kJ/mol)+2(−240kJ/mol)]−(−75kJ/mol)]\)
D \(\Delta H^0_{rxn}=[(−390kJ/mol)+(−240kJ/mol)]−(−75kJ/mol)]\)
▶️Answer/Explanation
Ans:C
The enthalpy change for a reaction can be estimated from \(\sum \Delta H^0_f,products-\sum \Delta H^0_f,reactants\) . Using the values given, \(\Delta H^0_{rxn}=[(−390kJ/mol)+2(−240kJ/mol)]−(−75kJ/mol)\).
Question
Refer to the following graph, which shows the heating curve for methane, \(CH_4\)
The enthalpy of vaporization of water is 40.7 kJ/mol. Which of the following best explains why the enthalpy of vaporization of methane is less than that of water?
(A) Methane does not exhibit hydrogen bonding, but water does.
(B) Methane has weaker dispersion forces.
(C) Methane has a smaller molar mass.
(D) Methane has a much lower density.
▶️Answer/Explanation
Ans:A
Question
Given the following reactions
\(Fe_2O_3 (s) + 3CO (s) \rightarrow 2Fe (s) + 3CO_2 (g)\) ∆H = -28.0 kJ
\(3Fe (s) + 4CO_2 (s) \rightarrow 4CO (g) + Fe_3O_4 (s)\) ∆H = +12.5 kJ
the enthalpy of the reaction of \(Fe_2O_3\) with CO
\(3Fe_2O_3 (s) + CO (g) \rightarrow CO_2 (g) + 2Fe_3O_4 (s)\) is ________ kJ.
A) 40.5 B) +109 C) -15.5 D) -109 E) -59.0
▶️Answer/Explanation
Ans: E
