If the heat of neutralization for a strong acid - base reaction is `-57.1 kJ`, what would be the heat released when `350 cm^(3)` at `0.20 M H_(2) SO_(4)` is mixed with `650 cm^(3)` of `0.10 M NaOH`?

Options

A. `37.1 kJ`
B. `3.71 kJ`
C. `3.17 kJ`
D. `0.317 kJ`

Answer

B

Explanation

According to the neutralization reaction,
`H^(+) (aq.) + OH^(-) (aq.) rarr H_(2) O (1) , Delta H^(@) = - 57.1 kJ mol^(-1)`
Thus, `57.1 kJ` energy is released for every 1 mol of water formed in the reaction.
millimoles of `H^(+) (aq.) = 2 ("millimoles of" H_(2) SO_(4))`
`= 2 (M_(H_(2)SO_(4)) V_(H_(2)SO_(4)))`
`= (0.20 xx 350) = 140 m mol`
millimole of `OH^(-) (aq.)` = millimoles of `NaOH`
`= M_(NaOH) V_(NaOH)`
`= (0.10) (650) = 65 mol`
This means `NaoH`is the limiting reactant.
`underset(140 m mol, 75 m mol)(H^(+) (aq.)) + underset(65 m mol, 0 m mol)(OH^(-) (aq.)) rarr underset(0 m mol, 65 m mol)(H_(2) O (1))`
Thus, 65 millimoles or `65 xx 10^(-3) mol` of `H_(2) O` are formed.
Hence, the amount of energy (heat) released will be
`(57.1 (kJ)/(mol)) (65 xx 10^(-3) mol)`
`= 3.71 kJ`