While performing a thermodynamics experiment,a student made the following observations:
$HCl + NaOH \rightarrow NaCl + H_{2}O$ $\Delta H = -57.3 \ kJ \ mol^{-1}$
$CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$ $\Delta H = -55.3 \ kJ \ mol^{-1}$
The enthalpy of ionization of $CH_{3}COOH$ as calculated by the student is $kJ \ mol^{-1}$. (nearest integer)
$HCl + NaOH \rightarrow NaCl + H_{2}O$ $\Delta H = -57.3 \ kJ \ mol^{-1}$
$CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$ $\Delta H = -55.3 \ kJ \ mol^{-1}$
The enthalpy of ionization of $CH_{3}COOH$ as calculated by the student is $kJ \ mol^{-1}$. (nearest integer)
- A$5$
- B$2$
- C$4$
- D$3$
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Similar Questions
Choose the reaction$(s)$ from the following options,for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$
$(1)$ $\frac{3}{2} O_{2(g)} \rightarrow O_{3(g)}$
$(2)$ $\frac{1}{8} S_{8(s)} + O_{2(g)} \rightarrow SO_{2(g)}$
$(3)$ $2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(l)}$
$(4)$ $2 C_{(g)} + 3 H_{2(g)} \rightarrow C_2H_{6(g)}$
MediumIIT 2019
View SolutionBased on the following thermochemical equations:
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.
$H_2O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)}; \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2}O_{2(g)} \to CO_{2(g)}; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(g)}; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = X \ kJ$
The value of $X$ is ...... $kJ$.
MediumAIPMT 1992
View SolutionThe heat of neutralization of four acids $A$,$B$,$C$,and $D$ are $-13.0$,$-12.6$,$-9.2$,and $-11.7 \ KCal/eq$ respectively when neutralized by $NaOH$. The order of acidic strength of the four acids will be:
Medium
View SolutionFrom the following bond energies:
$H-H$ bond energy $431.37 \text{ kJ mol}^{-1}$ $C=C$ bond energy $606.10 \text{ kJ mol}^{-1}$ $C-C$ bond energy $336.49 \text{ kJ mol}^{-1}$ $C-H$ bond energy $410.50 \text{ kJ mol}^{-1}$
Enthalpy for the reaction $CH_2=CH_2 + H-H \to CH_3-CH_3$ will be .............. $\text{kJ mol}^{-1}$
| $H-H$ bond energy | $431.37 \text{ kJ mol}^{-1}$ |
| $C=C$ bond energy | $606.10 \text{ kJ mol}^{-1}$ |
| $C-C$ bond energy | $336.49 \text{ kJ mol}^{-1}$ |
| $C-H$ bond energy | $410.50 \text{ kJ mol}^{-1}$ |
Enthalpy for the reaction $CH_2=CH_2 + H-H \to CH_3-CH_3$ will be .............. $\text{kJ mol}^{-1}$
MediumAIPMT 2009
View SolutionThe heat of combustion of solid benzoic acid at constant volume is $-321.30 \ kJ$ at $27^{\circ} C$. Its heat of combustion at constant pressure is :
Medium
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