The value of `DeltaE` for combustion of 16 g of `CH_4` is -885389 at 298 K. The `DeltaH` combustion for `CH_4` in `Jmol^(-1)` at this temperature will be : (Given that, `R = 8.314 JK^(-1) mol^(-1)`)

Options

A. `-55337`
B. `-880430`
C. `-885389`
D. `-890344`

Answer

D

Explanation

`CH_(4)(g) + 2O_(2) (g) to CO_(2)(g) + 2H_(2)O(l)Delta n_(g) = 1 - 3 = -2`
We know that , `Delta H = Delta E + Delta n_(g) RT`
`:. " "Delta H = (-885389)-(-2) xx 8.314 xx 298 = -885389 - 4955.1440 = -890344.4 J mol^(-1)`.