Enthalpy of neutralization of `HCl` with `NaOH` is `X`. The heat evolved when 500 ml of `2 N HCl` is mixed with 250 ml of `4 N NaOH` will be

Options

A. `500 X`
B. `100 X`
C. `X`
D. `10 X`

Answer

D

Explanation

The enthalpy of neutralization is described by the following equation:
`{:(H^(+) (aq.) +OH^(-) (aq.) rarr H_(2) O (1),,,Delta H = X):}`
Thus, heat released is `X` when 1 mol of `H_(2)O` is formed. For `HCl` as well as `NaOH`, normality `(N)` is equal to molarity `(M)`. Thus,
millimoles of `H^(+) (aq.) = M_(HCl) V_(HCl)`
`= (2) (500)`
`= 1000`
Moles of `H^(+) (aq.) = 1`
millimoles of `OH^(-) (aq.) = 1`
This implies that
`underset(0" mol")underset(1" mol")(H^(+)(aq.))+underset(0" mol")underset(1" mol")(OH^(-))rarrunderset(1" mol")underset(0" mol")(H_(2)O)`
Since 1 mol of `H_(2) O` is formed, heat released will be `X`.