Question

For the reaction at $300$ K

$A\left(g\right)+B\left(g\right)\to C\left(g\right)$

$\Delta H=-3.0kcal;\Delta S=-10.0cal/K$ value of $\Delta G$ is:

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

Answer 1

Answer: (B)

$-6600$cal

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

The change in Gibb's free energy is given by

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

whereas,

$\Delta H=$ Enthalpy of reaction

$\Delta S=$ Entropy of reaction $=-10cal/K$

As we know that,

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

whereas,

$\Delta U=$ Change in internal energy $=-3kcal=-3000cal$

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

$R=$ Gas constant $=2cal$

$T=300K$

$\therefore \Delta H=-3000+(-1\times 2\times 300)=-3600cal$

$\therefore \Delta G=-3600-(300\times (-10))=-3600-3000=-6000cal$

Hence the value of $\Delta G$ is $-6600cal$.

Hence, the correct option is $\text{B}$