Question

What is the total energy change for the following reaction:CO+H2O-CO2+H2

Answer 1

`\large \boxed{\text{-41.2 kJ/mol}}`

Step-by-step explanation:

Balanced equation: CO(g) + H₂O(g) ⟶ CO₂(g) + H₂(g)

We can calculate the enthalpy change of a reaction by using the enthalpies of formation of reactants and products

`\Delta_{\text{rxn}}H^(\circ) = \sum \left( \Delta_{\text{f}} H^(\circ) \text{products}\right) - \sum \left (\Delta_{\text{f}}H^(\circ) \text{reactants} \right)`

(a) Enthalpies of formation of reactants and products

`\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{CO(g)} & -110.5 \\\text{H$_(2)$O} & -241.8\\\text{CO$_(2)$(g)} & -393.5 \\\text{H$_(2)$(g)} & 0 \\\end{array}`

(b) Total enthalpies of reactants and products

`\begin{array}{ccr}\textbf{Substance} & \textbf{Contribution)/(kJ/mol})&\textbf{Sum} \\\text{CO(g)} & -110.5& -110.5 \\\text{H$_(2)$O(g)} &-241.8& -241.8\\\textbf{Total}&\textbf{for reactants} &\mathbf{ -352.3}\\&&\\\text{CO}_(2)(g) & -393.5&-393.5 \\\text{H}_(2) & 0 & 0\\\textbf{Total}&\textbf{for products} & \mathbf{-393.5}\end{array}`

(c) Enthalpy of reaction
`\Delta_{\text{rxn}}H^(\circ) = \sum \left( \Delta_{\text{f}} H^(\circ) \text{products}\right) - \sum \left (\Delta_{\text{f}}H^(\circ) \text{reactants} \right)= \text{-393.5 kJ/mol - (-352.3 kJ/mol}\\= \text{-393.5 kJ/mol + 352.3 kJ/mol} = \textbf{-41.2 kJ/mol}\\ \text{The total enthalpy change is $\large \boxed{\textbf{-41.2 kJ/mol}}$}`