Consider the reaction The temperature at which and is ............. K. (Round off to the Nearest Int — Chemical 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
💡 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**.