Question

The rate constants of a reaction at $300,K$ and $400,K$ are $1.5,s^{-1}$ and $3,s^{-1}$ respectively. The $E_a$ value is [$\log 2 = 0.3010,\ 2.303R=19,J,K^{-1}mol^{-1}$]

• A. $7.86,kJ$
• B. $68.6,kJ$
• C. $78.6,kJ$
• D. $6.86,kJ$
• E. $5.86,kJ$

Hint:

Chemistry Tip: If rate constant doubles with temperature increase, use $\log(k_2/k_1)=\log 2$ immediately to save time.

Answer 1

The Correct Option is D

Concept:
The dependence of rate constant on temperature is given by Arrhenius equation: $$k=Ae^{-E_a/RT}$$ For two temperatures: $$\log\left(\frac{k_2}{k_1}\right)=\frac{E_a}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$$ where:

• $k_1, k_2$ are rate constants
• $T_1, T_2$ are temperatures
• $E_a$ is activation energy

Step 1: Write the given data.
At $300K$: $$k_1=1.5,s^{-1}$$ At $400K$: $$k_2=3,s^{-1}$$ So: $$\frac{k_2}{k_1}=\frac{3}{1.5}=2$$ Hence: $$\log\left(\frac{k_2}{k_1}\right)=\log 2 = 0.3010$$
Step 2: Substitute into Arrhenius form.
$$0.3010=\frac{E_a}{19}\left(\frac{1}{300}-\frac{1}{400}\right)$$ Given: $$2.303R=19,J,K^{-1}mol^{-1}$$
Step 3: Simplify temperature term.
$$\frac{1}{300}-\frac{1}{400} =\frac{4-3}{1200} =\frac{1}{1200}$$ So: $$0.3010=\frac{E_a}{19}\times\frac{1}{1200}$$
Step 4: Solve for $E_a$.
$$E_a=0.3010\times19\times1200$$ $$E_a=6862.8,J,mol^{-1}$$ $$E_a\approx 6.86,kJ,mol^{-1}$$
Step 5: Final answer.
Therefore: $$\boxed{E_a=6.86,kJ,mol^{-1}}$$ Hence correct option is (D). :contentReference[oaicite:0]{index=0}