∆H for ${\rm{CaC}}{{\rm{O}}_3}\left( s \right)$$ \to {\rm{CaO}}\left( s \right) + {\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right)$ is 176 kJ$mo{l^{ - 1}}$ at 1240 K. The ∆U for the change is equal to:

NEET UG CHEMISTRY MCQ: THERMODYNAMICS

1 Medium

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Application +4 | -1

∆H for

\mathrm{CaC}{\mathrm{O}_3}\left( s \right)

\to \mathrm{CaO}\left( s \right) + \mathrm{C}{\mathrm{O}_2}\left( \mathrm{g} \right)

is 176 kJ

mo{l^{ - 1}}

at 1240 K. The ∆U for the change is equal to:

View answer and explanation

Correct option: B

165.6 kJ

The

\Delta U

for the change is 165.6 kJ, calculated using the formula

\Delta H = \Delta U + \Delta n_g RT

Practice All Chapter MCQs →

{T_i}

{T_f}

\mathrm{H}_2 \mathrm{O}(\mathrm{s}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})

\mathrm{X}_2 \mathrm{O_4(l}) \longrightarrow 2 \mathrm{XO}_2(\mathrm{g}), \Delta U=2.1 \mathrm{~kcal}, \Delta S=20 \mathrm{~cal} \mathrm{~K}^{-1}

300 \mathrm{~K}

\Delta G