JEE Chemistry 2026 Question with Solution

Given:

• $k_1 = 10^{6} e^{-\frac{30000}{T}}$
• $k_2 = 10^{4} e^{-\frac{24000}{T}}$
• $\ln 10 = 2.303$

Find: The temperature at which both rate constants become equal.

When the two Arrhenius rate constants are equal, set

$$10^{6} e^{-\frac{30000}{T}} = 10^{4} e^{-\frac{24000}{T}}$$

Taking natural logarithm on both sides,

$$\ln(10^{6}) - \frac{30000}{T} = \ln(10^{4}) - \frac{24000}{T}$$

Substitute $\ln 10 = 2.303$,

$$6(2.303) - \frac{30000}{T} = 4(2.303) - \frac{24000}{T}$$

Rearranging,

$$2(2.303) = \frac{6000}{T}$$

So,

$$T = \frac{6000}{4.606} \approx 1304 \; \text{K}$$

Therefore, the required temperature is $1304$.

Given: Two gaseous reactions with rate constants in Arrhenius form.

Find: Temperature at which $k_1 = k_2$.

Compare the logarithmic forms of both rate constants:

$$\ln k_1 = \ln\left(10^{6} e^{-\frac{30000}{T}}\right) = \ln(10^{6}) - \frac{30000}{T}$$ $$\ln k_2 = \ln\left(10^{4} e^{-\frac{24000}{T}}\right) = \ln(10^{4}) - \frac{24000}{T}$$

For equal rate constants,

$$\ln k_1 = \ln k_2$$

Hence,

$$\ln(10^{6}) - \frac{30000}{T} = \ln(10^{4}) - \frac{24000}{T}$$

Using $\ln(10^{n}) = n \ln 10$,

$$6 \ln 10 - \frac{30000}{T} = 4 \ln 10 - \frac{24000}{T}$$

Now substitute $\ln 10 = 2.303$,

$$6(2.303) - 4(2.303) = \frac{30000}{T} - \frac{24000}{T}$$ $$2(2.303) = \frac{6000}{T}$$ $$4.606 = \frac{6000}{T}$$ $$T = \frac{6000}{4.606} \approx 1304$$

Thus, the nearest integer value of temperature is $1304$.