The standard heats of formation of `CH_(4), H_(2)O` and `CH_(3)OH` are `-76, -242` and `-266` kJ/mole respectively. The enthalpy `CH_(3)OH_((l))+H_(2(g))rarrCH_(4(g))+H_(2)O_((l))`

Options

A. `-4` kJ/mole
B. `-556` kJ/mole
C. `-318` kJ/mole
D. `-52` kJ/mole

Answer

D

Explanation

`Delta H= Sigma Delta H_(f)^(@)` (products) `-Sigma Delta H_(f)^(@)` (reactants)
`Delta H= (-242.76) - (-266+ 0) = -52`kJ`