Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$
Assertion $(A)$: Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57 \ kJ \ mol^{-1}$.
Reason $(R)$: Enthalpy of neutralisation is the amount of heat liberated when one mole of $H^{+}$ ions furnished by acid combine with one mole of $OH^{-}$ ions furnished by base to form one mole of water.
In the light of the above statements,choose the correct answer from the options given below.
Assertion $(A)$: Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57 \ kJ \ mol^{-1}$.
Reason $(R)$: Enthalpy of neutralisation is the amount of heat liberated when one mole of $H^{+}$ ions furnished by acid combine with one mole of $OH^{-}$ ions furnished by base to form one mole of water.
In the light of the above statements,choose the correct answer from the options given below.
- A$(A)$ is true but $(R)$ is false.
- BBoth $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.
- C$(A)$ is false but $(R)$ is true.
- DBoth $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A)$.
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Similar Questions
The heats of formation of two compounds $X$ and $Y$ are $-84 \ kJ$ and $-156 \ kJ$ respectively. Which of the following statements is correct?
Easy
View SolutionGiven the reactions:
$C + \frac{1}{2}O_2 \to CO : \Delta H = -12 \ kJ$
$CO + \frac{1}{2}O_2 \to CO_2 : \Delta H = -10 \ kJ$
For the reaction $C + O_2 \to CO_2 : \Delta H = x \ kJ$,the value of $x$ is: (in $kJ$)
$C + \frac{1}{2}O_2 \to CO : \Delta H = -12 \ kJ$
$CO + \frac{1}{2}O_2 \to CO_2 : \Delta H = -10 \ kJ$
For the reaction $C + O_2 \to CO_2 : \Delta H = x \ kJ$,the value of $x$ is: (in $kJ$)
Medium
View SolutionIf enthalpies of formation of $C_2H_{4(g)}$,$CO_{2(g)}$ and $H_2O_{(l)}$ at $25 \ ^\circ C$ and $1 \ atm$ pressure are $52$,$-394$ and $-286 \ kJ \ mol^{-1}$ respectively,the enthalpy of combustion of $C_2H_{4(g)}$ will be.....$kJ \ mol^{-1}$.
MediumAIPMT 1995
View SolutionGiven the following thermochemical equations:
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)
$(i)$ $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}; \Delta H = -285 \text{ kJ}$
$(ii)$ $N_2O_{5(g)} + H_2O_{(l)} \rightarrow 2HNO_{3(l)}; \Delta H = -76.6 \text{ kJ}$
$(iii)$ $N_{2(g)} + 3O_{2(g)} + H_{2(g)} \rightarrow 2HNO_{3(l)}; \Delta H = -348.2 \text{ kJ}$
Calculate the $\Delta H$ for the reaction: $2N_{2(g)} + 5O_{2(g)} \rightarrow 2N_2O_{5(g)}$. (in $\text{ kJ}$)
The molar neutralization heat for $KOH$ and $HNO_3$ as compared to the molar neutralization heat of $NaOH$ and $HCl$ is:
Easy
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