The decomposition of phosphine `4PH_3 (g) to P_4 (g) + 6H_2 (g)` has the 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)` . What is the value of rate constant at 310 K? [R = `8.314 JK ^(-1) mol^(-1)`]

Explanation

According to Arrhenius equation,
` log "" (k_2)/(k_1) = (E_a)/( 2.303 R) [(T_2-T_1)/(T_1 T_2)]`
here ` k_1= 6.0 xx 10^(-4) s^(-1) " " T_1 = 300 K k_2 =?`
` E_a = 3.05 xx 10^(-5) J : R = 8.314 JK^(-1) mol^(-1)`
` therefore log "" (k_2)/(k_1) = ( 3.05 xx 10^5 )/(2.303 xx 8.314 )=[(310-300 )/(310 xx 300 )] = 1.7128`
` (k_2 )/(k_1) = 51.62 or k_2 =k_1 xx 51 .62 = 6.0 xx 10^(-4) xx 51 .62 s^(-1)`
`= 3.1 xx 10^(-2) s^(-1)`