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)

• A. \(x + y + RT\)
• B. \(x - y + RT\)
• C. \(x + y + 2RT\)
• D. \(x - y + 2RT\)

Hint:

Remember that \(\Delta H\) accounts for both internal energy change and pressure-volume work at constant pressure.

Answer 1

The Correct Option is C

Step 1: Connect enthalpy and internal energy.
Enthalpy change relates to internal energy change by $\Delta_r H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ is the change in moles of gas.
Step 2: Apply the first law.
The first law says $\Delta U = q + w$, taking heat added as positive and work done by the system as negative.
Step 3: Insert the given symbols.
Heat absorbed is $q = +x$ and work done by the system is $w = -y$, so \[ \Delta U = x - y \]
Step 4: Count the gas moles.
The reaction $A(g) \rightarrow B(g) + C(g)$ goes from $1$ mole of gas to $2$ moles, so $\Delta n_g = 2 - 1 = 1$.
Step 5: Assemble the enthalpy.
\[ \Delta_r H = (x - y) + (1)RT = x - y + RT \]
Step 6: Match the option.
This is option 3 in the listed set, $x - y + RT$.
\[ \boxed{x - y + RT} \]