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. \( N_{2}H_{4(g)} + H_{2(g)} \rightarrow 2NH_{3(g)} \); \( \Delta_{r}H^{\circ}_{2} = -187.6\text{ kJ} \)}
Show 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.
Step 1: Understanding the Question:
We need to find the enthalpy change for the target reaction: \( N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \) using Hess's Law. Step 2: Key Formula or Approach:
Hess's Law of Constant Heat Summation states that the total enthalpy change for a reaction is the same regardless of whether the reaction occurs in one step or several steps. Step 3: Detailed Explanation:
Let's add the two given equations:
Equation (i): \( 2H_{2(g)} + N_{2(g)} \rightarrow N_{2}H_{4(g)} \)
Equation (ii): \( N_{2}H_{4(g)} + H_{2(g)} \rightarrow 2NH_{3(g)} \)
Adding (i) and (ii):
\( (2H_{2} + N_{2} + N_{2}H_{4} + H_{2})_{(g)} \rightarrow (N_{2}H_{4} + 2NH_{3})_{(g)} \)
Cancelling \( N_{2}H_{4(g)} \) from both sides and combining \( H_{2} \) terms:
\( N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)} \)
The resulting enthalpy change \( \Delta_{r}H^{\circ} \) will be the sum of individual enthalpies:
\[ \Delta_{r}H^{\circ} = \Delta_{r}H^{\circ}_{1} + \Delta_{r}H^{\circ}_{2} \]
\[ \Delta_{r}H^{\circ} = 95.4\text{ kJ} + (-187.6\text{ kJ}) \]
\[ \Delta_{r}H^{\circ} = -92.2\text{ kJ} \] Step 4: Final Answer:
The standard enthalpy change for the synthesis of ammonia is -92.2 kJ.