For the following reaction at 300 K: \[ \text{A}_2(g) + 3\text{B}_2(g) \rightarrow 2\text{AB}_3(g) \] The enthalpy change is \(+15 \, \text{kJ}\), then the internal energy change is:
- 19988.4 J
- 200 J
- 1999 J
- 1.9988 kJ
The Correct Option is A
Solution and Explanation
To find the internal energy change (\( \Delta U \)) for the given reaction, we will use the relation between enthalpy change (\( \Delta H \)) and internal energy change (\( \Delta U \)). This relationship includes the work done due to expansion or compression under constant pressure:
\(\Delta H = \Delta U + \Delta nRT\)
Where:
- \(\Delta H\) is the enthalpy change (given as \( +15 \, \text{kJ} \)).
- \(\Delta U\) is the internal energy change.
- \(\Delta n\) is the change in moles of gas during the reaction.
- \(R\) is the universal gas constant (\( 8.314 \, \text{J/mol K} \)).
- \(T\) is the temperature in Kelvin (given as \( 300 \, \text{K} \)).
First, we calculate the change in moles of gas (\( \Delta n \)). For the given reaction:
- Reactants: \( 1 \, \text{mol of } \text{A}_2(g) \) and \( 3 \, \text{mol of } \text{B}_2(g) \) for a total of \( 4 \) moles.
- Products: \( 2 \, \text{mol of } \text{AB}_3(g) \).
Change in moles of gas, \( \Delta n = \text{moles of products} - \text{moles of reactants} = 2 - 4 = -2 \).
Given \(\Delta H = +15 \, \text{kJ} = 15000 \, \text{J}\) (since 1 kJ = 1000 J).
Substitute into the formula:
\(\Delta U = \Delta H - \Delta nRT\)
Substitute the known values:
\(\Delta U = 15000 \, \text{J} - (-2 \times 8.314 \, \text{J/mol K} \times 300 \, \text{K}) = 15000 \, \text{J} + 2 \times 8.314 \times 300\)
\(\Delta U = 15000 \, \text{J} + 4988.4 \, \text{J}\)
\(\Delta U = 19988.4 \, \text{J}\)
Therefore, the internal energy change (\( \Delta U \)) for this reaction is 19988.4 J.
Thus, the correct answer is
19988.4 J
.
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