Question

The amount of heat evolved when one mole of ${\text{H}}_2{\text{SO}}_4$ reacts with two moles of $\text{NaOH}$ is:

• A. $13.7\text{k cal}$
  
• B. More than $13.7\text{k cal}$
  
• C. Less than $13.7\text{k cal}$
  
• D. Cannot say

Answer 1

The correct option is B More than $13.7\text{k cal}$


${\text{H}}_2{\text{SO}}_4+2\text{NaOH}\to {\text{Na}}_2{\text{SO}}_4+2{\text{H}}_2\text{O}$;

One mole of $H_{2}S{O}_4$ is completely neutralised by the two moles of $NaOH$.

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

$\Rightarrow \text{Gram equivalents of}{H}_{2}S{O}_4=1\times 2=2$
$\text{Heat evolved in the neutralisation of}1\text{gm eq. of strong acid}=-13.7\text{k cal}$

$\Rightarrow \text{Heat evolved in the neutralisation of}2\text{gm eq. of strong acid}=-13.7\times 2\text{k cal}$
$\Rightarrow \Delta H=2\times (-13.7)=-27.4\text{k cal}$