Question:
At T(K) in a reaction \(A(g) \rightarrow B(g) + C(g)\), x J of heat was absorbed and y J of work is done by the system. What is \(\Delta_r H\) (in J) for the reaction? (R= gas constant)
At T(K) in a reaction \(A(g) \rightarrow B(g) + C(g)\), x J of heat was absorbed and y J of work is done by the system. What is \(\Delta_r H\) (in J) for the reaction? (R= gas constant)
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Remember that \(\Delta H\) accounts for both internal energy change and pressure-volume work at constant pressure.
Updated On: Jun 9, 2026
- \(x + y + RT\)
- \(x - y + RT\)
- \(x + y + 2RT\)
- \(x - y + 2RT\)
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The Correct Option is C
Solution and Explanation
Concept:
Enthalpy change \(\Delta H\) is related to internal energy change \(\Delta U\) by the relation \(\Delta H = \Delta U + \Delta n_g RT\).
Step 1: Use the first law of thermodynamics to find \(\Delta U\).
From the first law, \(\Delta U = q + w\). Given heat absorbed \(q = +x\) and work done by the system \(w = -y\): \[ \Delta U = x - y \]
Step 2: Determine the change in moles of gas (\(\Delta n_g\)).
The reaction is \(A(g) \rightarrow B(g) + C(g)\). \[ \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = (1 + 1) - 1 = 1 \]
Step 3: Calculate \(\Delta_r H\).
\[ \Delta_r H = \Delta U + \Delta n_g RT \] \[ \Delta_r H = (x - y) + (1)RT = x - y + RT \] \[ \boxed{x - y + RT} \]
Step 1: Use the first law of thermodynamics to find \(\Delta U\).
From the first law, \(\Delta U = q + w\). Given heat absorbed \(q = +x\) and work done by the system \(w = -y\): \[ \Delta U = x - y \]
Step 2: Determine the change in moles of gas (\(\Delta n_g\)).
The reaction is \(A(g) \rightarrow B(g) + C(g)\). \[ \Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants}) = (1 + 1) - 1 = 1 \]
Step 3: Calculate \(\Delta_r H\).
\[ \Delta_r H = \Delta U + \Delta n_g RT \] \[ \Delta_r H = (x - y) + (1)RT = x - y + RT \] \[ \boxed{x - y + RT} \]
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