If ${CH}_{3} COOH + {OH}^{-} \to {CH}_{3} {COO}^{-} + H_{2} O + q_{1}$ and $H^{+} + {OH}^{-} ⟶ H_{2} O + q_{2}$ then the enthalpy change for the reaction ${CH}_{3} COOH ⟶ {CH}_{3} {COO}^{-} + H^{+}$ is equal to:

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$q_{1} + q_{2}$

$q_{1} - q_{2}$

$q_{2} - q_{1}$

$- q_{1} - q_{2}$

answer is C.

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Detailed Solution

1) ${CH}_{3} COOH + {OH}^{-} \to {CHCOO}^{-} + H_{2} OΔH = - q_{1}$

$H^{+} + {OH}^{-} \to H_{2} OΔH = - q_{2}$

2) $H_{2} O \to H^{+} + {OH}^{-} ΔH = + q_{2}$

Adding 1 and 2

${CH}_{3} COOH \to {CHCOO}^{-} + H^{+} ΔH = - q_{1} + q_{2}$

Enthalpy change for reaction is a result. $= q_{2} - q_{1}$

Hence, option C is correct.

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