Question

Calculate the standard enthalpy change for synthesis of ammonia gas from following data. i. \( 2H_{2(g)} + N_{2(g)} \rightarrow N_{2}H_{4(g)} \); \( \Delta_{r}H^{\circ}_{1} = 95.4\text{ kJ} \) ii. \( N_{2}H_{4(g)} + H_{2(g)} \rightarrow 2NH_{3(g)} \); \( \Delta_{r}H^{\circ}_{2} = -187.6\text{ kJ} \)}

• A. -92.2 kJ
• B. -46.1 kJ
• C. -138.3 kJ
• D. -283.2 kJ

Hint:

Hess's Law allows you to add enthalpies of intermediate steps. Always check if the question asks for the total reaction or per mole of product.

Answer 1

The Correct Option is B

Step 1: Target Reaction
The standard synthesis of ammonia is: $\frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)} \rightarrow NH_{3(g)}$.
Step 2: Add Equations
Adding equation (i) and (ii):
$(2H_2 + N_2) + (N_2H_4 + H_2) \rightarrow N_2H_4 + 2NH_3$
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
$\Delta H_{total} = \Delta_r H^\circ_1 + \Delta_r H^\circ_2 = 95.4 + (-187.6) = -92.2$ kJ.
Step 3: Adjust for 1 Mole
The value $-92.2$ kJ is for 2 moles of $NH_3$. For the standard enthalpy of synthesis (per mole):
$\Delta_f H^\circ = \frac{-92.2}{2} = -46.1$ kJ.
Step 4: Conclusion
Hence, the standard enthalpy change is $-46.1$ kJ.
Final Answer:(B)