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Hess' Law

The enthalpie...

Question

The enthalpies of neutralisation of $N a O H$ and ${N H}_{4} O H$ by $H C l$ are $- 13680$ calories and $- 12270 c a l$ respectively. What would be the enthalpy change of one gram equivalent of $N a O H$ is added to one gram equivalent of ${N H}_{4} C l$ is solution? Assume that ${N H}_{4} O H$ and $N a C l$ are quantitatively obtained (only magnitude in cal).

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Solution

Enthalpy of neutralization is change in enthaply when 1 g equivalent of an acid reacts completely with 1 gram equivalent of base in dilute solution.
$R_{1} : N a O H (a q .) + H C l (a q) \to N a C l (a q) + H_{2} O; Δ H_{1} = - 13680 c a l$
$R_{2} : N H_{4} O H (a q .) + H C l (a q) \to N H_{4} C l (a q) + H_{2} O; Δ H_{2} = - 12270 c a l$
$R_{1} - R_{2} \Rightarrow N H_{4} C l (a q .) + N a O H (a q) \to N H_{4} O H l (a q) + N a C (a q .)$
Given that 1 gram equivalent of $N a O H$ is added to 1 gram equivalent of $N H_{4} C l$
$\Rightarrow Δ H = Δ H_{1} - Δ H_{2} = - 13680 + 12270 = - 1410 c a l$
Answer = 1410

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