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The decomposition of phosphine, 4PH(3)...

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))`

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