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(A) The synthesis of ammonia gas is represented by the equation:$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$To obtain this equation,we add the

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$-92.2$

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(A) The synthesis of ammonia gas is represented by the equation:$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$To obtain this equation,we add the two given reactions:Reaction $i$: $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$Reaction $ii$: $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$Adding reaction $i$ and reaction $ii$:$(2 H_{2(g)} + N_{2(g)}) + (N_{2}H_{4(g)} + H_{2(g)}) \longrightarrow N_{2}H_{4(g)} + 2 NH_{3(g)}$$N_{2(g)} + 3 H_{2(g)} \longrightarrow 2 NH_{3(g)}$The total enthalpy change is $\Delta_{r}H^{0} = \Delta_{r}H_{1}^{0} + \Delta_{r}H_{2}^{0}$$\Delta_{r}H^{0} = 95.4 \ kJ + (-187.6 \ kJ) = -92.2 \ kJ$This is the enthalpy change for the production of $2 \ moles$ of $NH_{3}$.Therefore,the correct option is $A$.

Calculate the standard enthalpy change for the synthesis of ammonia gas from the following data:
$i$. $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$
$ii$. $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$ (in $kJ$)

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Calculate the standard enthalpy change for the synthesis of ammonia gas from the following data:$i$. $2 H_{2(g)} + N_{2(g)} \longrightarrow N_{2}H_{4(g)}$; $\Delta_{r}H_{1}^{0} = 95.4 \ kJ$$ii$. $N_{2}H_{4(g)} + H_{2(g)} \longrightarrow 2 NH_{3(g)}$; $\Delta_{r}H_{2}^{0} = -187.6 \ kJ$ (in $kJ$)