Question

For the reaction at $300K$
$A\left(g\right)+B\left(g\right)\to C\left(g\right)$
$\Delta E=-3.0Kcal;\Delta S=-10.0cal/K$ value of $\Delta G$ is:

• A. $-600cal$
• B. $6600cal$
• C. $-6000cal$
• D. None

Answer 1

The correct option is A $-600cal$

Given:-

$\Delta E=-3Kcal$

$\Delta S=-10cal/K$

$T=300K$

$A_{\left(g\right)}+B_{\left(g\right)}⟶C_{\left(g\right)}$

For the above reaction,

$\Delta n_g=n_P-n_R=1-(1+1)=-1$

As we know that,

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

$\Rightarrow \Delta H=-3000+(-1\times 2\times 300)$

$\Rightarrow \Delta H=-3000-600=-3600cal$

Now again as we know that,

$\Delta G=\Delta H-T\Delta S$

$\Rightarrow \Delta G=(-3600)-(300\times (-10))$

$\Rightarrow \Delta G=-3600+3000$

$\Rightarrow \Delta G=-600cal$