Question

$1$ mole of $H_{2}S{O}_4$ is mixed with $2$ moles of NaOH. The heat evolved will be:

• A. $57.3$ kJ
• B. $2\times 57.3$ kJ
• C. $57.3/2$ kJ
• D. Cannot be predicted

Answer 1

The correct option is B $2\times 57.3$ kJ

$H_{2}S{O}_4+2NaOH⟶{Na}_{2}S{O}_4+2H_{2}O$

According to reaction,

$1$ mole $H_{2}S{O}_4$ completely neutralised by $2$ mole of $NaOH$.

As we know,

Gram equivalent $=$ no. of moles $\times$ Valency factor

Valency factor of $H_{2}S{O}_4=2$

Therefore,

Gram equivalent of $H_{2}S{O}_4=1\times 2=2$

As we know,

Heat evolved in the neutralisation of $1$ gm eq. of strong acid $=57.3kJ$

Heat evolved in neutralisation of $2$ gm eq. of strong acid $=(57.3\times 2)kJ$

Hence the heat evolved will be $2\times 57.3kJ$.

Hence, the correct option is $\text{B}$