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Chapter 6: Problem 85

Water gas is produced from the reaction of steam with coal: $$ \mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{CO}(g) $$ Assuming that coal is pure graphite, calculate \(\Delta H^{\circ}\) for this reaction.

Short Answer

Expert verified
The standard enthalpy change for the reaction between graphite and steam to produce water gas is 131.3 kJ/mol. This is calculated using the formula ∆H° = sum of the standard enthalpies of formation of the products − sum of the standard enthalpies of formation of the reactants, and using the given standard enthalpies of formation for CO, H2O, C (graphite), and H2.

Step by step solution

01

1. Gather the relevant data

We need to gather the standard enthalpies of formation for the substances involved in the reaction. For this, we can consult a table of standard enthalpies of formation. We will find the following values: - ∆Hf° (CO) = -110.5 kJ/mol - ∆Hf° (H2O) = -241.8 kJ/mol - ∆Hf° (C, graphite) = 0 kJ/mol (since graphite is the standard state of carbon) - ∆Hf° (H2) = 0 kJ/mol (since H2 is in its standard state)
02

2. Use the formula to calculate ∆H°

Now that we have the necessary data, we can apply the formula: ∆H° = [∆Hf° (H2) + ∆Hf° (CO)] - [∆Hf° (C, graphite) + ∆Hf° (H2O)] Plugging in the values: ∆H° = [0 + (-110.5)] - [0 + (-241.8)] Calculating the sum: ∆H° = -110.5 + 241.8 ∆H° = 131.3 kJ/mol The standard enthalpy change for the reaction between graphite and steam to produce water gas is 131.3 kJ/mol.

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