Standard molar enthalpy of formation,$\Delta _{f}H^{o}$ is just a special case of enthalpy of reaction,$\Delta _{r}H^{o}$. Is the $\Delta _{r}H^{o}$ for the following reaction same as $\Delta _{f}H^{o}$? Give reason for your answer. $CaO_{(s)} + CO_{2_{(g)}} \to CaCO_{3_{(s)}}$; $\Delta _{r}H^{o} = -178.3 \ kJ \ mol^{-1}$
(N/A) No,the $\Delta _{r}H^{o}$ for the given reaction is not the same as $\Delta _{f}H^{o}$.
The standard molar enthalpy of formation,$\Delta _{f}H^{o}$,is defined as the standard enthalpy change for the formation of $1 \ mol$ of a compound from its constituent elements in their most stable reference states.
The reaction for the formation of $CaCO_{3(s)}$ is:
$Ca_{(s)} + C_{(s)} + \frac{3}{2} O_{2(g)} \to CaCO_{3(s)}$
In the given reaction,$CaO_{(s)} + CO_{2(g)} \to CaCO_{3(s)}$,the product is formed from compounds ($CaO$ and $CO_{2}$) rather than from elements in their standard states.
Therefore,$\Delta _{r}H^{o} \neq \Delta _{f}H^{o}$.
The standard molar enthalpy of formation,$\Delta _{f}H^{o}$,is defined as the standard enthalpy change for the formation of $1 \ mol$ of a compound from its constituent elements in their most stable reference states.
The reaction for the formation of $CaCO_{3(s)}$ is:
$Ca_{(s)} + C_{(s)} + \frac{3}{2} O_{2(g)} \to CaCO_{3(s)}$
In the given reaction,$CaO_{(s)} + CO_{2(g)} \to CaCO_{3(s)}$,the product is formed from compounds ($CaO$ and $CO_{2}$) rather than from elements in their standard states.
Therefore,$\Delta _{r}H^{o} \neq \Delta _{f}H^{o}$.
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Similar Questions
The enthalpy changes for the following reactions are given:
$Cl_{2(g)} = 2Cl_{(g)}, 242.3 \, kJ \, mol^{-1}$; $I_{2(g)} = 2I_{(g)}, 151.0 \, kJ \, mol^{-1}$
$ICl_{(g)} = I_{(g)} + Cl_{(g)}, 211.3 \, kJ \, mol^{-1}$; $I_{2(s)} = I_{2(g)}, 62.76 \, kJ \, mol^{-1}$
Given that the standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is $...... \, kJ \, mol^{-1}$.
$Cl_{2(g)} = 2Cl_{(g)}, 242.3 \, kJ \, mol^{-1}$; $I_{2(g)} = 2I_{(g)}, 151.0 \, kJ \, mol^{-1}$
$ICl_{(g)} = I_{(g)} + Cl_{(g)}, 211.3 \, kJ \, mol^{-1}$; $I_{2(s)} = I_{2(g)}, 62.76 \, kJ \, mol^{-1}$
Given that the standard states of iodine and chlorine are $I_{2(s)}$ and $Cl_{2(g)}$,the standard enthalpy of formation for $ICl_{(g)}$ is $...... \, kJ \, mol^{-1}$.
Difficult
View SolutionFind the value of $Q$ from the following equations:
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
$(i)$ $C_{(s)} + O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = Q \ kJ$
$(ii)$ $C_{(s)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{(g)}$ $\Delta H = -x \ kJ$
$(iii)$ $CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \longrightarrow CO_{2_{(g)}}$ $\Delta H = -y \ kJ$
The equation $\frac{1}{2} H_2 + \frac{1}{2} Cl_2 \to HCl$ $(\Delta H_{298} = -22.060 \ kcal)$ means:
Medium
View SolutionIf $C$ (diamond) $\rightarrow C$ (graphite) $+ X \ kJ \ mol^{-1}$,$C$ (diamond) $+ O_{2(g)} \rightarrow CO_{2(g)} + Y \ kJ \ mol^{-1}$,and $C$ (graphite) $+ O_{2(g)} \rightarrow CO_{2(g)} + Z \ kJ \ mol^{-1}$,at constant temperature,then which of the following relations is correct?
On the basis of the thermochemical equations:
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.
$H_{2}O_{(g)} + C_{(s)} \to CO_{(g)} + H_{2(g)} \quad \Delta H = 131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)} \quad \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_{2}O_{(g)} \quad \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \to CO_{2(g)} \quad \Delta H = X \ kJ$
The value of $X$ will be $.... \ kJ$.
Medium
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