Question

For the reaction; ${FeCO}_{3\left(s\right)}⟶Fe{O}_{\left(s\right)}+C{O}_2\left(g\right)$, $\Delta H=82.8kJ$ at ${25}^{0}C$. What is $\Delta U$ at ${25}^{0}C$?

• A. $82.8kJ$
• B. $80.32kJ$
• C. $-2394.77kJ$
• D. $85.28kJ$

Answer 1

Answer: (B)

$80.32kJ$

Solution:- (B) $80.32kJ$

${FeC{O}_3}_{\left(s\right)}⟶{FeO}_{\left(s\right)}+{C{O}_2}_{\left(g\right)}$

From first law of thermodynamics-

$\Delta U=\Delta H-\Delta nRT$

$\Delta H=82.8KJ$

$T=25℃=(25+273)K=298K$

$R=8.314J{mol}^{-1}{K}^{-1}=8.314\times {10}^{-3}kJ{mol}^{-1}{K}^{-1}$

$\Delta n=n_p-n_r=\left(1\right)-0=1$

$\therefore \Delta U=82.8-(1\times 8.314\times {10}^{-3}\times 298)$

$\Rightarrow \Delta U=82.8-2.48=80.32kJ$

Hence for th given reaction, $\Delta U$ at $25℃$ is $80.32kJ$.