Question

What will be the standard internal energy change for the reaction at $298 \,K$?
$OF_{2\left(g\right)}+H_{2}O_{\left(g\right)} \to O_{2\left(g\right)}+2HF_{\left(g\right)} ; \Delta H^{\circ} = -310\,kj$

• A. $-312.5 \,kj$
• B. $-125.03 \,kj$
• C. $-310 \,kj$
• D. $-156 \,kj$

Answer 1

Answer: (A)

$\Delta H = \Delta U + \Delta n_{g}RT$
$\Delta H = -310 \times 10^{3}\,J$,
$\Delta n_{g} = 3-2=1$,
$R= 8.314\,J\,K^{-1}\,mol^{-1}$,
$T = 298\,K$
$\Delta U= -310 \times 10^{-3}-\left(1\times8.314 \times298\right)$
$= -312.477 \times 10^3 \,J = -312.5 \,kJ$