The first order rate constant for the decomposition of CaCO at 700 K is 6.36 × 10 s and activation e — Chemical Kinetics Chemistry Question
Question
The first order rate constant for the decomposition of CaCO at 700 K is 6.36 × 10 s and activation energy is 209 kJ mol . Its rate constant (in s ) at 600 K is x × 10 . The value of x is _______.? (Nearest integer) [Given R = 8.31 J K mol ; log 6.36 × 10 = –2.19, 10 = 1.62 × 10 ] 3 –3 ֪–1 –1 –1 –6 –1 –1 –3 –4.79 –5
💡 Solution & Explanation
# Solution: Finding Rate Constant at Different Temperature **Step 1: Identify the Formula** Use the Arrhenius equation in logarithmic form: $$\log\frac{k_2}{k_1} = \frac{E_a}{2.303R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$ **Step 2: Organize Given Data** - k₁ = 6.36 × 10⁻³ s⁻¹ at T₁ = 700 K - k₂ = ? at T₂ = 600 K - Eₐ = 209 kJ/mol = 209,000 J/mol - R = 8.31 J K⁻¹ mol⁻¹ **Step 3: Calculate Temperature Term** $$\frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{700} - \frac{1}{600} = 0.001429 - 0.001667 = -0.000238 \text{ K}^{-1}$$ **Step 4: Calculate the Logarithmic Ratio** $$\log\frac{k_2}{k_1} = \frac{209,000}{2.303 × 8.31} × (-0.000238) = \frac{209,000}{19.11} × (-0.000238)$$ $$= 10,941 × (-0.000238) = -2.60$$ **Step 5: Find k₂** $$\log k_2 - \log(6.36 × 10^{-3}) = -2.60$$ $$\log k_2 - (-2.19) = -2.60$$ $$\log k_2 = -4.79$$ $$k_2 = 10^{-4.79} = 1.62 × 10^{-5} \text{ s}^{-1}$$ **Step 6: Express as x ×