Question

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:

• A. 19988.4 J
• B. 200 J
• C. 1999 J
• D. 1.9988 kJ

Answer 1

Answer: (A)

19988.4 J

Answer 2

The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is given by:
$\Delta H = \Delta U + \Delta n_g RT$
where:
$\Delta n_g$ is the change in the number of moles of gas.
R is the gas constant (8.314 J K$^{-1}$ mol$^{-1}$).
T is the temperature in Kelvin.
For the given reaction:
$\Delta n_g$ = moles of gaseous products – moles of gaseous reactants
$\Delta n_g = 2 - 4 = -2$
$\Delta H = +15$ kJ = $15 \times 10^3$ J
T = 300 K
$\Delta U = \Delta H - \Delta n_g RT = 15000 J - (-2 mol)(8.314 J K^{-1}mol^{-1})(300 K)$
$\Delta U = 15000 + 4988.4 = 19988.4 J$