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Consider the following reactions:
(i) $H^+(aq) + OH^-(aq) = H_2O(l)$, $\Delta H = -X_1$ kJ mol$^{-1}$
(ii) $H_2(g) + \dfrac{1}{2}O_2(g) = H_2O(l)$, $\Delta H = -X_2$ kJ mol$^{-1}$
(iii) $CO_2(g) + H_2(g) = CO(g) + H_2O(l)$, $\Delta H = -X_3$ kJ mol$^{-1}$
(iv) $C_2H_2(g) + \dfrac{5}{2}O_2(g) = 2CO_2(g) + H_2O(l)$, $\Delta H = -X_4$ kJ mol$^{-1}$
Enthalpy of formation of $H_2O(l)$ is
(i) $H^+(aq) + OH^-(aq) = H_2O(l)$, $\Delta H = -X_1$ kJ mol$^{-1}$
(ii) $H_2(g) + \dfrac{1}{2}O_2(g) = H_2O(l)$, $\Delta H = -X_2$ kJ mol$^{-1}$
(iii) $CO_2(g) + H_2(g) = CO(g) + H_2O(l)$, $\Delta H = -X_3$ kJ mol$^{-1}$
(iv) $C_2H_2(g) + \dfrac{5}{2}O_2(g) = 2CO_2(g) + H_2O(l)$, $\Delta H = -X_4$ kJ mol$^{-1}$
Enthalpy of formation of $H_2O(l)$ is
A
$+X_3$ kJ mol$^{-1}$
B
$-X_4$ kJ mol$^{-1}$
C
$+X_1$ kJ mol$^{-1}$
D
$-X_2$ kJ mol$^{-1}$
Detailed Solution
The enthalpy of formation is the heat evolved or absorbed when one mole of a substance is formed from its constituent elements in their standard states.
The elements of water in their standard states are $H_2(g)$ and $O_2(g)$.
Reaction (ii), $H_2(g) + \dfrac{1}{2}O_2(g) = H_2O(l)$, shows the formation of one mole of $H_2O(l)$ from its elements, so $-X_2$ represents the enthalpy of formation of $H_2O(l)$.
Reaction (i) is the neutralisation of $H^+$ and $OH^-$ ions, so $-X_1$ is the enthalpy of neutralisation.
Reaction (iii) forms water from a compound ($CO_2$), not from the elements alone.
Reaction (iv) is the combustion of acetylene, so $-X_4$ is its enthalpy of combustion.
So the enthalpy of formation of $H_2O(l)$ is $-X_2$ kJ mol$^{-1}$.
The elements of water in their standard states are $H_2(g)$ and $O_2(g)$.
Reaction (ii), $H_2(g) + \dfrac{1}{2}O_2(g) = H_2O(l)$, shows the formation of one mole of $H_2O(l)$ from its elements, so $-X_2$ represents the enthalpy of formation of $H_2O(l)$.
Reaction (i) is the neutralisation of $H^+$ and $OH^-$ ions, so $-X_1$ is the enthalpy of neutralisation.
Reaction (iii) forms water from a compound ($CO_2$), not from the elements alone.
Reaction (iv) is the combustion of acetylene, so $-X_4$ is its enthalpy of combustion.
So the enthalpy of formation of $H_2O(l)$ is $-X_2$ kJ mol$^{-1}$.