The decomposition of phosphine,
`4PH_(3)(g) to P_(4)(g) + 6H_(2)(g)` has rate law , Rate = `k[PH_(3)]`. The rate constant is `6.0 xx 10^(-4) s^(-1)` at 300 K and activation energy is `3.05 xx 10^(5)` J `mol^(-1)`. Calculate the value of the rate constant at 310 K (`R= 8.314 JK^(-1)mol^(-1))`

Explanation

To calculate the rate constant \( k_2 \) at 310 K for the decomposition of phosphine, we will use the Arrhenius equation in the logarithmic form. The steps are as follows:

### Step 1: Write down the Arrhenius equation in logarithmic form
The Arrhenius equation relates the rate constants at two different temperatures as follows:

\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
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