The formation of ammonia is shown by the reaction $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}, \Delta_r H^o = -91.8 \, kJ \, mol^{-1}$. What will be the enthalpy of reaction for the decomposition of $NH_3$ according to the reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}; \Delta_r H^o = ?$ ..... $kJ \, mol^{-1}$
- A$-91.8$
- B$+91.8$
- C$-45.9$
- D$+45.9$
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Similar Questions
The standard enthalpy of formation $(\Delta_f H^{\circ})$ of ammonia is $-46.2 \ kJ \ mol^{-1}$. What is the $\Delta_r H^{\circ}$ of the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
When equal volumes of $1 \ M \ HCl$ and $1 \ M \ H_2SO_4$ are separately neutralized by an excess volume of $1 \ M \ NaOH$ solution,$x \ J$ and $y \ J$ of heat are liberated,respectively. The value of $y / x$ is . . . .
Consider the following data:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by:
$\Delta_{f}H^{\Theta}(CH_{4}, g) = -X \ kJ \ mol^{-1}$
Enthalpy of sublimation of graphite = $Y \ kJ \ mol^{-1}$
Dissociation enthalpy of $H_{2} = Z \ kJ \ mol^{-1}$
The bond enthalpy of $C-H$ bond is given by:
DifficultJEE MAIN 2026
View SolutionUse the data from the table to estimate the enthalpy of formation of $CH_3CHO$.
Bond Bond Enthalpy $(kJ \ mol^{-1})$ Enthalpy of formation $(kJ \ mol^{-1})$ $C-H$ $400$ $C(g): 700$ $C-C$ $350$ $H(g): 200$ $C=O$ $700$ $O(g): 250$
| Bond | Bond Enthalpy $(kJ \ mol^{-1})$ | Enthalpy of formation $(kJ \ mol^{-1})$ |
|---|---|---|
| $C-H$ | $400$ | $C(g): 700$ |
| $C-C$ | $350$ | $H(g): 200$ |
| $C=O$ | $700$ | $O(g): 250$ |
The heats of combustion of carbon and carbon monoxide are $-394 \, kJ \, mol^{-1}$ and $-285 \, kJ \, mol^{-1}$ respectively. Find the heat of formation of $CO$ in $kJ \, mol^{-1}$.
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