Question

The temperature at which the rate constants of the given below two gaseous reactions become equal is ____________ K (Nearest integer). 
\[ X \longrightarrow Y, \qquad k_1 = 10^{6} e^{-\frac{30000}{T}} \] \[ P \longrightarrow Q, \qquad k_2 = 10^{4} e^{-\frac{24000}{T}} \] Given: \( \ln 10 = 2.303 \)

Hint:

When two Arrhenius rate constants are equal, equate their logarithmic forms to eliminate the exponential term.

Answer 1

Correct Answer: 1304

Step 1: Set the two rate constants equal.
\[ 10^{6} e^{-\frac{30000}{T}} = 10^{4} e^{-\frac{24000}{T}} \] Step 2: Take natural logarithm on both sides.
\[ \ln(10^{6}) - \frac{30000}{T} = \ln(10^{4}) - \frac{24000}{T} \] Step 3: Substitute \( \ln 10 = 2.303 \).
\[ 6(2.303) - \frac{30000}{T} = 4(2.303) - \frac{24000}{T} \] Step 4: Simplify and solve for \( T \).
\[ 2(2.303) = \frac{6000}{T} \] \[ T = \frac{6000}{4.606} \approx 1304\ \text{K} \] Final Answer:
\[ \boxed{1304} \]