Given the reaction
3 NO2(g) + H2O(l) �¨ 2 HNO3(l) + NO(g) ƒ¢rH = -72.0 kJ,
calculate the molar enthalpy of reaction, rH for:
1) NO2(g)
2) H20 (l)
3) HNO3 (l)
4) NO (g)
Express you answer in Kj/mol
To calculate the molar enthalpy of reaction (ΔH) for each substance involved in the reaction, we need to use the given reaction equation and the information provided.
1) ΔH(NO2):
In the given reaction equation, the coefficient of NO2 is 3. This means that for every 3 moles of NO2 consumed, the reaction releases -72.0 kJ of energy. Therefore, the molar enthalpy of reaction for NO2 is:
ΔH(NO2) = -72.0 kJ / 3 mol = -24.0 kJ/mol
2) ΔH(H2O):
In the given reaction equation, there is no specific coefficient for H2O. However, since H2O is on the reactant side, it is being consumed in the reaction. The coefficient of NO2 is 3, so that means 3 moles of H2O are consumed. The energy released for this reaction corresponds to -72.0 kJ. Therefore, the molar enthalpy of reaction for H2O is:
ΔH(H2O) = -72.0 kJ / 3 mol = -24.0 kJ/mol
3) ΔH(HNO3):
In the given reaction equation, the coefficient of HNO3 is 2. This means that for every 2 moles of HNO3 produced, the reaction releases -72.0 kJ of energy. Therefore, the molar enthalpy of reaction for HNO3 is:
ΔH(HNO3) = -72.0 kJ / 2 mol = -36.0 kJ/mol
4) ΔH(NO):
In the given reaction equation, the coefficient of NO is 1. This means that for every 1 mole of NO produced, the reaction releases -72.0 kJ of energy. Therefore, the molar enthalpy of reaction for NO is:
ΔH(NO) = -72.0 kJ / 1 mol = -72.0 kJ/mol
Therefore, the molar enthalpies of reaction are:
1) ΔH(NO2) = -24.0 kJ/mol
2) ΔH(H2O) = -24.0 kJ/mol
3) ΔH(HNO3) = -36.0 kJ/mol
4) ΔH(NO) = -72.0 kJ/mol