For the reaction, αA + bB → cC + dD, the plot of log k vs is given below: The temperature at which t — Chemical Kinetics Chemistry Question

💡 Solution & Explanation

**Step 1: Identify the relationship from Arrhenius equation** The Arrhenius equation in logarithmic form is: $$\log k = \log A - \frac{E_a}{2.303RT}$$ This shows that log k vs 1/T is linear with slope = -E_a/2.303R **Step 2: Determine the slope from the graph** From the given plot of log k vs 1/T: - At one point: 1/T = 2.0 × 10⁻³ K⁻¹, log k = -4 - At another point: 1/T = 1.8 × 10⁻³ K⁻¹, log k = -5 Slope = (-5 - (-4))/(1.8 - 2.0) × 10⁻³ = -1/(-0.2 × 10⁻³) = 5000 K **Step 3: Calculate activation energy** Slope = -E_a/2.303R 5000 = E_a/(2.303 × 8.314) E_a = 5000 × 19.14 = 95,700 J/mol ≈ 95.7 kJ/mol **Step 4: Use the given condition at 500 K** At T₁ = 500 K, k₁ = 10 s⁻¹ (given: log k₁ = 1) At T₂ = ?, k₂ = 10 s⁻¹ (same rate constant) Wait—using two-temperature form: $$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ At 500 K: log k = 1, so this is the reference point. **Step 5: Find temperature where log k = 1** From graph, when log k = 1, reading 1/T