Chapter 20: Problem 15

Hydrogen is produced commercially by the reaction of methane with steam: $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ a. Calculate $\Delta H^{\circ}$ and $\Delta S^{\circ}$ for this reaction (use the data in Appendix 4$)$ b. What temperatures will favor product formation at standard conditions? Assume $\Delta H^{\circ}$ and $\Delta S^{\circ}$ do not depend on temperature.

Short Answer

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a. The standard enthalpy change (∆H°) for the reaction is 205.1 kJ/mol, and the standard entropy change (∆S°) is 314 J/mol⋅K. b. The reaction will favor product formation at standard conditions at temperatures above approximately 653 K.

Step by step solution

1. Find standard enthalpy and entropy of individual compounds

To find the standard enthalpies and entropies for the reaction, first look up the values from Appendix 4 for each compound involved in the reaction. We list the values for methane (CH4), water (H2O), carbon monoxide (CO), and hydrogen (H2). Standard Enthalpy of Formation (∆H°[f]): CH4(g): -74.8 kJ/mol H2O(g): -241.8 kJ/mol CO(g): -110.5 kJ/mol H2(g): 0 kJ/mol Standard Entropies (S°): CH4(g): 186.3 J/mol⋅K H2O(g): 188.8 J/mol⋅K CO(g): 197.7 J/mol⋅K H2(g): 130.7 J/mol⋅K

2. Calculate standard enthalpy and entropy change

To calculate the standard enthalpy change (∆H°) and standard entropy change (∆S°) for this reaction, use the formula: ∆H° = Σ[(moles × ∆H°[f]_products) - (moles × ∆H°[f]_reactants)] ∆S° = Σ[(moles × S°_products) - (moles × S°_reactants)] For this reaction, ∆H° = (1 × (-110.5) + 3 × (0)) - (1 × (-74.8) + 1 × (-241.8)) ∆H° = 205.1 kJ/mol ∆S° = (1 × 197.7 + 3 × 130.7) - (1 × 186.3 + 1 × 188.8) ∆S° = 314 J/mol⋅K b. What temperatures will favor product formation at standard conditions?

3. Determine the temperature that favors product formation

To find the temperature that favors product formation, we need to determine when the Gibbs free energy change (∆G°) is negative (reaction is spontaneous). Use the equation: ∆G° = ∆H° - T∆S° Rearrange to find the temperature where ∆G° = 0 (the reaction is spontaneous at higher temperatures): T = (∆H°) / (∆S°) T = (205.1 kJ/mol) / (0.314 kJ/mol⋅K) T ≈ 653 K The temperature at which the reaction will favor product formation at standard conditions is approximately 653 K. At temperatures above this value, the reaction will be spontaneous, favoring the formation of products.

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Short Answer

Step by step solution

1. Find standard enthalpy and entropy of individual compounds

2. Calculate standard enthalpy and entropy change

3. Determine the temperature that favors product formation

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