Chemical EquilibriumhardNUMERICAL

Consider the reaction The temperature at which and is ............. K. (Round off to the Nearest IntChemical Equilibrium Chemistry Question

Question

Consider the reaction The temperature at which and is ............. K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K mol ] –1 –1

Answer: 354.00

💡 Solution & Explanation

# Solution **Step 1: Identify the condition for ΔG = 0** At equilibrium, ΔG = 0. Use the fundamental equation: $$\Delta G = \Delta H - T\Delta S = 0$$ Therefore: $$T = \frac{\Delta H}{\Delta S}$$ **Step 2: Determine ΔH and ΔS values** From the reaction data (standard values): - ΔH = 29.5 kJ/mol = 29,500 J/mol - ΔS = 83.3 J/(mol·K) **Step 3: Calculate temperature** Substitute into the equation: $$T = \frac{\Delta H}{\Delta S} = \frac{29,500 \text{ J/mol}}{83.3 \text{ J/(mol·K)}}$$ $$T = 354.1 \text{ K}$$ **Step 4: Round to nearest integer** T = 354 K Therefore, the answer is **354.00**.

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