# For the reaction:$X_{2}O_{4}\left ( l \right )\rightarrow 2XO_{2}(g)$$\Delta U=2.1k \ cal, \Delta S=20 cal\ K^{-1}$  at 300KHence $\Delta G$ is :- Option 1) 2.7k cal Option 2) -2.7 k cal Option 3) 9.3k cal Option 4) -9.3k cal

P Plabita

Molar heat capacity for isobaric process C(p) -

$dH=n\: C_{P}\: dT$

- wherein

$dH=dE+d(PV)$

or

$dH=dE+n\, R\, dT$

$\Delta H= \Delta U + \Delta n_{g}RT$

Given, $\Delta U=2.1 Kcal\;\;\;\; \Delta n_{g}=2$

$R- 2 \times 10^{-3}Kcal;\;\;T=300 K$

$\therefore \Delta H= 2.1+2 \times 2 \times 10^{-3} \times 300=3.3 Kcal$

Also, $\Delta G= \Delta H-T \Delta S$

Given, $\Delta S=20 \times 10^{-3}Kcal\;K^{-1}$

On putting the value of $\Delta H$ and $\Delta S$ in the equation, we get -

$\Delta G = 3.3-300 \times 20 \times 10^{-3}= -2.7 Kcal$

Option 1)

2.7k cal

This option is incorrect.

Option 2)

-2.7 k cal

This option is correct.

Option 3)

9.3k cal

This option is incorrect.

Option 4)

-9.3k cal

This option is incorrect.

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