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Use the following heat of formation table to answer the questions below.



The Standard Enthalpy and Entropy of Various Substances




Substance		$\Delta H_f^{\circ}(kJ / mol)$$S^{\circ}(J / K \cdot mol)$

$C_4 H_{10(g)}$		-126310

$CaC_{2(s)}$		-6370.

$Ca(OH)_{2(s)}$		-98783

$C_2 H_{2(g)}$		227201

$CO_{2(g)}$		-394214

$H_{2(g)}$		0131

$H_2 O_{(g)}$		-242189

$H_2 O_{(l)}$		-28670.

$NH_{3(g)}$		-46193

$NO_{(g)}$		90.211

$NO_{2(g)}$		34240.

$N_2 O_{(g)}$		82220.

$O_{2(g)}$		0205

$O_{3(g)}$		143239





  1. Using data from the heat of formation table above, calculate the enthalpy of reaction for:

$3 H_{2(g)} + O_{3(g)} \rightarrow 3 H_2 O_{(g)}$


  1. Using data from the heat of formation table above, calculate the heat of reaction for:

$2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$


  1. Using data from the heat of formation table above, calculate the heat of reaction for:

$N_2 O_{(g)} + NO_{2(g)} \rightarrow 3 NO_{(g)}$

Verified Answer :

  • Use the formula \(\Delta H_{rxn}^{\circ} = \sum n_p \Delta H_f^{\circ}(\text{products}) - \sum n_r \Delta H_f^{\circ}(\text{reactants})\) for each reaction.
  • For \(3H_{2(g)} + O_{3(g)} \rightarrow 3H_2O_{(g)}\), the result is \(3(-242) - (3(0) + 143) = -869 \text{ kJ/mol}\).
  • For \(2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}\), the result is \(2(34) - (2(90) + 0) = -112 \text{ kJ/mol}\).
  • For \(N_2O_{(g)} + NO_{2(g)} \rightarrow 3NO_{(g)}\), the result is \(3(90) - (82 + 34) = 154 \text{ kJ/mol}\).
  • Final results: \(\boxed{-869 \text{ kJ/mol}, -112 \text{ kJ/mol}, 154 \text{ kJ/mol}}\).

Explanation

  1. Analysis of Data and Problem Setup

We analyze the provided chemical reactions to determine their standard enthalpies of reaction (\(\Delta H_{rxn}^{\circ}\)). The standard enthalpy of reaction is calculated using the standard enthalpies of formation (\(\Delta H_f^{\circ}\)) of the reactants and products involved:
\[\Delta H_{rxn}^{\circ} = \sum n_p \Delta H_f^{\circ}(\text{products}) - \sum n_r \Delta H_f^{\circ}(\text{reactants})\]
where \(n_p\) and \(n_r\) are the stoichiometric coefficients for products and reactants, respectively. The required standard enthalpies of formation for the chemical species involved are taken from the provided table:

  • \(\Delta H_f^{\circ}(H_{2(g)}) = 0 \text{ kJ/mol}\)
  • \(\Delta H_f^{\circ}(O_{3(g)}) = 143 \text{ kJ/mol}\)
  • \(\Delta H_f^{\circ}(H_2O_{(g)}) = -242 \text{ kJ/mol}\)
  • \(\Delta H_f^{\circ}(NO_{(g)}) = 90 \text{ kJ/mol}\)
  • \(\Delta H_f^{\circ}(O_{2(g)}) = 0 \text{ kJ/mol}\)
  • \(\Delta H_f^{\circ}(NO_{2(g)}) = 34 \text{ kJ/mol}\)
  • \(\Delta H_f^{\circ}(N_2O_{(g)}) = 82 \text{ kJ/mol}\)
  1. Calculating Enthalpy for the First Reaction

For the reaction \(3H_{2(g)} + O_{3(g)} \rightarrow 3H_2O_{(g)}\):
The enthalpy of reaction is calculated as:
\[\Delta H_{rxn}^{\circ} = [3 \cdot \Delta H_f^{\circ}(H_2O_{(g)})] - [3 \cdot \Delta H_f^{\circ}(H_{2(g)}) + 1 \cdot \Delta H_f^{\circ}(O_{3(g)})]\]
Substituting the known values:
\[\Delta H_{rxn}^{\circ} = [3 \cdot (-242 \text{ kJ/mol})] - [3 \cdot 0 \text{ kJ/mol} + 1 \cdot 143 \text{ kJ/mol}]\]
\[\Delta H_{rxn}^{\circ} = -726 \text{ kJ/mol} - 143 \text{ kJ/mol} = -869 \text{ kJ/mol}\]
Thus, the standard enthalpy of reaction is \(-869 \text{ kJ/mol}\).

  1. Calculating Enthalpy for the Second Reaction

For the reaction \(2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}\):
The enthalpy of reaction is calculated as:
\[\Delta H_{rxn}^{\circ} = [2 \cdot \Delta H_f^{\circ}(NO_{2(g)})] - [2 \cdot \Delta H_f^{\circ}(NO_{(g)}) + 1 \cdot \Delta H_f^{\circ}(O_{2(g)})]\]
Substituting the known values:
\[\Delta H_{rxn}^{\circ} = [2 \cdot 34 \text{ kJ/mol}] - [2 \cdot 90 \text{ kJ/mol} + 1 \cdot 0 \text{ kJ/mol}]\]
\[\Delta H_{rxn}^{\circ} = 68 \text{ kJ/mol} - 180 \text{ kJ/mol} = -112 \text{ kJ/mol}\]
Thus, the standard enthalpy of reaction is \(-112 \text{ kJ/mol}\).

  1. Calculating Enthalpy for the Third Reaction

For the reaction \(N_2O_{(g)} + NO_{2(g)} \rightarrow 3NO_{(g)}\):
The enthalpy of reaction is calculated as:
\[\Delta H_{rxn}^{\circ} = [3 \cdot \Delta H_f^{\circ}(NO_{(g)})] - [1 \cdot \Delta H_f^{\circ}(N_2O_{(g)}) + 1 \cdot \Delta H_f^{\circ}(NO_{2(g)})]\]
Substituting the known values:
\[\Delta H_{rxn}^{\circ} = [3 \cdot 90 \text{ kJ/mol}] - [1 \cdot 82 \text{ kJ/mol} + 1 \cdot 34 \text{ kJ/mol}]\]
\[\Delta H_{rxn}^{\circ} = 270 \text{ kJ/mol} - (82 + 34) \text{ kJ/mol} = 270 \text{ kJ/mol} - 116 \text{ kJ/mol} = 154 \text{ kJ/mol}\]
Thus, the standard enthalpy of reaction is \(154 \text{ kJ/mol}\).

  1. Conclusion and Summary

In summary, the standard enthalpies of reaction for the three given equations are:

  1. \(3H_{2(g)} + O_{3(g)} \rightarrow 3H_2O_{(g)}\) is \(-869 \text{ kJ/mol}\).
  2. \(2NO_{(g)} + O_{2(g)} \rightarrow 2NO_{2(g)}\) is \(-112 \text{ kJ/mol}\).
  3. \(N_2O_{(g)} + NO_{2(g)} \rightarrow 3NO_{(g)}\) is \(154 \text{ kJ/mol}\).

The final answers are \(\boxed{-869 \text{ kJ/mol}}\), \(\boxed{-112 \text{ kJ/mol}}\), and \(\boxed{154 \text{ kJ/mol}}\).

Examples

Calculating the enthalpy of a reaction is a fundamental concept in chemical engineering and industrial chemistry. For example, when designing a fuel cell, engineers calculate the enthalpy change of the combustion reaction to estimate how much heat energy will be released during the process. This helps in managing thermal regulation within the system, ensuring that it operates efficiently and safely without overheating. Understanding these thermodynamic properties allows for the precise control of energy conversion in power plants, industrial reactors, and even portable batteries.

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