Question

Find temperature (in Kelvin) at which rate constant are equal for the following reaction? 
\(\text{A $\rightarrow$ B, K = 10$^4$ e$^{-24000/T}$} \)
\(\text{P $\rightarrow$ Q, K = 10$^6$ e$^{-30000/T}$} \)

Hint:

When comparing rate constants for reactions at different temperatures, use the Arrhenius equation to solve for the temperature where the rate constants become equal.

Answer 1

Correct Answer: 1303

Step 1: Setting up the equation.
At the temperature where the rate constants are equal, we set the two expressions for K equal to each other: \[ 10^4 e^{-24000/T} = 10^6 e^{-30000/T} \]
Step 2: Simplifying the equation.
Taking the natural logarithm of both sides: \[ \ln (10^4) - \frac{24000}{T} = \ln (10^6) - \frac{30000}{T} \] \[ 4 \ln 10 - \frac{24000}{T} = 6 \ln 10 - \frac{30000}{T} \] \text{Since} \(\ln 10 = 2.3026\), we substitute: \[ 4(2.3026) - \frac{24000}{T} = 6(2.3026) - \frac{30000}{T} \] \[ 9.2104 - \frac{24000}{T} = 13.8156 - \frac{30000}{T} \]
Step 3: Solving for T.
Now, solving for T: \[ 9.2104 - 13.8156 = \frac{30000}{T} - \frac{24000}{T} \] \[ -4.6052 = \frac{6000}{T} \] \[ T = \frac{6000}{4.6052} = 1303 \, \text{K} \]
Step 4: Conclusion.
Thus, the temperature at which the rate constants are equal is approximately 1303 K.