| For the reaction,\[{{X}_{2}}{{O}_{4}}(l)\to 2X{{O}_{2}}(g)\] |
| \[\Delta U=2.1\,kcal,\,\,\Delta S=20\,cal\,\,{{K}^{-1}}\] at 300 K. |
| Hence, \[\Delta G\] is [AIPMT 2014] |
A) 2.7 kcal
B) -2.7 kcal
C) 9.3 kcal
D) -9.3 kcal
Correct Answer: B
Solution :
| The change in Gibbs free energy is given by |
| \[\Delta G=\Delta H-T\Delta S\] |
| where, \[\Delta H\] = enthalpy of the reaction |
| \[\Delta S\] = entropy of the reaction |
| Thus, in order to determine\[\Delta G\], the values of \[\Delta H\]must be known, the value of \[\Delta H\]can be calculated by the equation |
| \[\Delta H=\Delta U+\Delta {{n}_{g}}RT\] |
| where \[\Delta U\] = change in internal energy |
| \[\Delta {{n}_{g}}\] = (number of moles of gaseous products) - (number of moles of gaseous reactants) |
| = 2 - 0 = 2 |
| R = gas constant = 2 cal |
| But, \[\Delta H=\Delta u+\Delta {{n}_{g}}RT\] |
| \[\Delta u=2.1kcal=2.1\times {{10}^{3}}cal\] |
| \[[\because \,1kcal\,={{10}^{3}}cal]\] |
| \[\therefore \] \[\Delta H=(2.1\times {{10}^{3}})+(2\times 2\times 300)=3300cal\] |
| Hence, \[\Delta G=\Delta H-T\Delta S\] |
| Þ \[\Delta G=(3300)-(300-20)\] |
| \[\Delta G=-2700\,cal\] |
| \[\therefore \] \[\Delta G=-2.7kcal\] |
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