The value of $\Delta_{f} H^{o}$ for $NH_3$ is $-91.8 \ kJ \ mol^{-1}$. Calculate the enthalpy change for the following reaction: $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$
The formation reaction for $1 \ mole$ of $NH_3$ is: $\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \to NH_{3(g)}$; $\Delta_f H^o = -91.8 \ kJ \ mol^{-1}$.
For the formation of $2 \ moles$ of $NH_3$,the reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta_r H^o = 2 \times (-91.8 \ kJ \ mol^{-1}) = -183.6 \ kJ \ mol^{-1}$.
The given reaction is the reverse of the formation of $2 \ moles$ of $NH_3$.
Therefore,for the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the enthalpy change is: $\Delta_r H^o = -(-183.6 \ kJ \ mol^{-1}) = +183.6 \ kJ \ mol^{-1}$.
For the formation of $2 \ moles$ of $NH_3$,the reaction is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$; $\Delta_r H^o = 2 \times (-91.8 \ kJ \ mol^{-1}) = -183.6 \ kJ \ mol^{-1}$.
The given reaction is the reverse of the formation of $2 \ moles$ of $NH_3$.
Therefore,for the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$,the enthalpy change is: $\Delta_r H^o = -(-183.6 \ kJ \ mol^{-1}) = +183.6 \ kJ \ mol^{-1}$.
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Similar Questions
Average bond enthalpy of water is $464.5 \text{ kJ mol}^{-1}$. If the energy required to break the first $O-H$ bond is $502 \text{ kJ mol}^{-1}$,how much energy per mol is required to break the second $O-H$ bond?
Three thermochemical equations are given below:
$(i) C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\circ = x \ kJ \ mol^{-1}$
$(ii) C_{(graphite)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\circ = y \ kJ \ mol^{-1}$
$(iii) CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\circ = z \ kJ \ mol^{-1}$
Based on the above equations,find out which of the relationship given below is correct.
$(i) C_{(graphite)} + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\circ = x \ kJ \ mol^{-1}$
$(ii) C_{(graphite)} + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\circ = y \ kJ \ mol^{-1}$
$(iii) CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\circ = z \ kJ \ mol^{-1}$
Based on the above equations,find out which of the relationship given below is correct.
MediumNEET 2013
View SolutionGiven:
$(i) \, C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = x \, kJ \, mol^{-1}$
$(ii) \, C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\ominus = y \, kJ \, mol^{-1}$
$(iii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = z \, kJ \, mol^{-1}$
Based on the above thermochemical equations,find out which one of the following algebraic relationships is correct?
$(i) \, C(\text{graphite}) + O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = x \, kJ \, mol^{-1}$
$(ii) \, C(\text{graphite}) + \frac{1}{2} O_{2(g)} \to CO_{(g)}; \Delta_r H^\ominus = y \, kJ \, mol^{-1}$
$(iii) \, CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}; \Delta_r H^\ominus = z \, kJ \, mol^{-1}$
Based on the above thermochemical equations,find out which one of the following algebraic relationships is correct?
DifficultJEE MAIN 2019
View SolutionHow much heat in $kJ$ is produced by the combustion of $15.5 \, g$ of propane? ${C_3H_8} + 5{O_2} \to 3{CO_2} + 4{H_2O}; \Delta{H^o} = -2219 \, kJ/mol$
Medium
View SolutionIn order to decompose $9 \ g$ of water,$142.5 \ kJ$ of heat is required. Hence,the enthalpy of formation of water is ...... $kJ/mol$.
Medium
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