# Question 57ac0

Mar 28, 2017

Remember

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

Thus $\Delta r G = \Delta r H - T \Delta r S$ because Change in a system is proportional to overall enthalpy, entropy or Gibbs free energy.

Thus for the reaction

$C O C {l}_{2} \rightarrow C O + C {l}_{2}$

We know everything except $\Delta r S$ so take it as x. Now write down the equation

("-206 kJ")/"mol "= "-220kJ"/"mol" - 298.15K xx x

Solve for $x$

$298.15 \times x = \left(\text{-220 kJ/mol") - ("-206 kJ/mol}\right)$

$298.15 \times x = - 14$

$x = - \frac{14}{298.15 K}$

= $\Delta r S = \text{-046.95623kJ"/"mol}$

As we are assuming that the $\Delta r S$ which we have found and $\Delta r H$ remains constant at 450K too we can now set up the equation

DeltarG^@ = "-220kJ"/"mol" - (450K xx "-046.95623kJ/mol")

DeltarG^@ = "-220kJ"/"mol" - (450K xx "-046.95623kJ/mol")#

$\Delta r {G}^{\circ} = \text{-220 kJ"/"mol} - \left(- 21.130 .3035\right)$

$\Delta r {G}^{\circ} = \text{-220kJ"/"mol} + 21.1303035$

$\Delta r {G}^{\circ} = \text{-198.8696965kJ"/"mol}$

Now we know that entropy or enthalpy cannot remain constant at different temerature so

$\Delta r {G}^{\circ} \approx \text{-198.8696965kJ"/"mol}$