The amount of heat evolved when one mole of $H_{2} {SO}_{4}$ reacts with two moles of $NaOH$ is:

13.7 k cal

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More than

13.7 k cal

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Less than

13.7 k cal

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Solution

The correct option is B More than $13.7 k cal$

$H_{2} {SO}_{4} + 2 NaOH \to {Na}_{2} {SO}_{4} + 2 H_{2} O$ ;

One mole of $H_{2} S O_{4}$ is completely neutralised by the two moles of $N a O H$ .

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

$\Rightarrow Gram equivalents of H_{2} S O_{4} = 1 \times 2 = 2$
$Heat evolved in the neutralisation of 1 gm eq. of strong acid = - 13.7 k cal$

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

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The amount of heat evolved when one mole of H2SO4 reacts with two moles of NaOH is:

The amount of heat evolved when one mole of $H_{2} {SO}_{4}$ reacts with two moles of $NaOH$ is: