Enthalpy of `CH_(4)+(1)/(2)O_(2)rarrCH_(3)OH` is
negative. If enthalpy of combustion of `CH_(4)` and `CH_(3)OH` are `x` and `y` respectively, then which relation is correct?

Options

A. `X gt Y`
B. `X lt Y`
C. `X = Y`
D. `X ge Y`

Answer

A

Explanation

If all thermochemical data to be used in predicting reaction enthalpy for a desired equation are heats of combustion, Hess's law can be expressed as
`Delta_(r) H^(@) = sum_(i) a_(i) Delta_(c ) H^(@)` (products) `+ sum_(i) b_(i) Delta_(c ) H^(@)` (reactants)
`= - (- Y) + (-X)`
`= Y - X`
Since `Y - X` is negative, this implies that `X gt Y`. Alternatively,
(a) `CH_(4) + O_(2) rarr CO_(2) + 2 H_(2) O, Delta H_(1) = - X kJ`
(b) `CH_(3) OH + (3)/(2) O_(2) rarr CO_(2) + 2 H_(2) O, Delta_(2) = - Y kJ`
Subtracting (b) and from (a), we get
`CH_(4) (g) + (1)/(2) O_(2) (g) rarr CH_(3) OH, Delta H_(3) =` negative that is,
`(-X) - (-Y)` is negative
Or `Y - X` is negative which is possible only when
`X gt Y`