NEET Solved Paper 2014 Question 14
Question: For the reaction, $ X _2O _4(l)\to 2XO _2(g) $ $ \Delta U=2.1kcal,\Delta S=20cal,{K^{-1}} $ at $ 300K. $ Hence, $ \Delta G $ is [AIPMT 2014]
Options:
A) 2.7 kcal
B) -2.7 kcal
C) 9.3 kcal
D) -9.3 kcal
Show Answer
Answer:
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 $
