Correct Answer - Option 1 : 166

**Concept:**

**Arrhenius equation:**

Arrhenius equation gives the dependence of the rate constant of a chemical reaction on the absolute temperature, a pre-exponential factor and other constants of the reaction.

\(k={{e}^{\frac{-{{E}_{a}}}{RT}}}\) ----(1)

Where,

k is the rate constant,

T is the absolute temperature (in Kelvin),

E_{a} is the activation energy for the reaction (in the same units as RT),

R is the universal gas constant.

**Calculation:**

We can get the activation energy using Arrhenius equation which is given by the formula:

\(k={{e}^{\frac{-{{E}_{a}}}{RT}}}\)

From the question, the reaction is:

H_{2} (g) + I_{2} (g) → 2HI (g)

Now, taking logarithm on both sides on equation (1),

\(\Rightarrow \ln k=-\frac{{{E}_{a}}}{RT}\)

Now, the s for the given reaction is:

\(\log \frac{{{k}_{2}}}{{{k}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{1}{{{T}_{1}}}-\frac{1}{{{T}_{2}}} \right]\)

T_{1 }= 327°C + 273 = 600 K

T_{2} = 527°C + 273 = 800 K

\(\Rightarrow \log \left( \frac{1}{2.5\times {{10}^{-4}}} \right)=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{1}{600}-\frac{1}{800} \right]\)

\(\Rightarrow \log \left( \frac{1}{2.5\times {{10}^{-4}}} \right)=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{4-3}{2400} \right]\)

\(\Rightarrow \log \left( \frac{1}{2.5\times {{10}^{-4}}} \right)=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{1}{2400} \right]\)

\(\Rightarrow 3.602=\frac{{{E}_{a}}}{2.303\times 8.314}\left[ \frac{1}{2400} \right]\)

⇒ E_{a }= 2.303 × 8.314 × 2400 × 3.602

∴

** E**_{a }= 165.5 KJ.mol^{-1}