Given $\Delta H_{f}{ }^{\circ}$ for $CO _{2}(g), CO (g)$ and $H _{2} O (g)$ are $ -393.5, -110.5 \text { and } -241.8 kJ mol ^{-1}$, respectively. The $\Delta H_{f}^{\circ}$ [in $kJ mol ^{-1}$ ] for the reaction $CO _{2}(g)+ H _{2}(g) \longrightarrow CO (g)+ H _{2} O (g)$ is
- 524.1
- -262.5
- -41.7
- 41.2
The Correct Option is D
Solution and Explanation
$\Delta H_{f}^{\circ}=(\Delta H)_{\text {products }}-(\Delta H)_{\text {reactants }}$
Hence, $\Delta H_{f}{ }^{\circ}=\left[\Delta H_{f_{ CO }}+\Delta H_{f_{ H _{2} O }}\right]-\left[\Delta H_{f CO _{2}}\right] \ldots( i )$
Given, $\Delta H_{f CO _{2}(s)}^{\circ}=-393.5 kJ mol ^{-1}$
$\Delta H_{f CO }^{\circ}=-110.5 \,kJ mol ^{-1}$
$\Delta H_{f H _{2} O }^{\circ}=-241.8 kJ mol ^{-1}$
Put there value in equation
$\Delta H_{f^{\circ}}=-110.5-241.8+393.5=41.2$
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Concepts Used:
Enthalpy change
Enthalpy Change refers to the difference between the heat content of the initial and final state of the reaction. Change in enthalpy can prove to be of great importance to find whether the reaction is exothermic or endothermic.
Formula for change in enthalpy is:-
dH = dU + d(PV)
The above equation can be written in the terms of initial and final states of the system which is defined below:
UF – UI = qP –p(VF – VI)
Or qP = (UF + pVF) – (UI + pVI)
Enthalpy (H) can be written as H= U + PV. Putting the value in the above equation, we obtained:
qP = HF – HI = ∆H
Hence, change in enthalpy ∆H = qP, referred to as the heat consumed at a constant pressure by the system. At constant pressure, we can also write,
∆H = ∆U + p∆V
Standard Enthalpy of Reaction
To specify the standard enthalpy of any reaction, it is calculated when all the components participating in the reaction i.e., the reactants and the products are in their standard form. Therefore the standard enthalpy of reaction is the enthalpy change that occurs in a system when a matter is transformed by a chemical reaction under standard conditions.
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