For strong acid and strong base neutralisation,the net chemical change is $H^{+} + OH^{-} \longrightarrow H_2O_{(l)}$; $\Delta_r H^{\circ} = -55.84 \ kJ \ mol^{-1}$. If the enthalpy of neutralisation of $CH_3COOH$ by $NaOH$ is $-49.86 \ kJ \ mol^{-1}$,then the enthalpy of ionisation of $CH_3COOH$ is:
- A$5.98 \ kJ \ mol^{-1}$
- B$-5.98 \ kJ \ mol^{-1}$
- C$105.7 \ kJ \ mol^{-1}$
- D$-59.8 \ kJ \ mol^{-1}$
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Similar Questions
Find the value of $x$ in $\text{kJ}$ using the following equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} : \Delta H = 131 \ \text{kJ}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} : \Delta H = -282 \ \text{kJ}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} : \Delta H = -242 \ \text{kJ}$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} : \Delta H = x \ \text{kJ}$
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} : \Delta H = 131 \ \text{kJ}$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} : \Delta H = -282 \ \text{kJ}$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(g)} : \Delta H = -242 \ \text{kJ}$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} : \Delta H = x \ \text{kJ}$
Medium
View SolutionAt constant pressure,the heat of formation of a compound is not dependent on temperature,when
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View SolutionAt $1 \ bar$ and $298 \ K$,the standard molar enthalpy of formation of which substance is zero?
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View SolutionEnthalpies of formation of $CCl_{4(g)}$,$H_2O_{(g)}$,$CO_{2(g)}$ and $HCl_{(g)}$ are $-105$,$-242$,$-394$ and $-92 \ kJ \ mol^{-1}$ respectively. The magnitude of enthalpy of the reaction given below is $...... \ kJ \ mol^{-1}$ (nearest integer): $CCl_{4(g)} + 2H_2O_{(g)} \rightarrow CO_{2(g)} + 4HCl_{(g)}$
State Hess's Law of Constant Heat Summation.
Easy
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