For the reaction $N H_{4} C l (S) \to N H_{3} (g) + H C l (g)$ at $25^{0}$ C, enthalpy change $△ = + 177 k J m o l^{- 1}$ and entropy change $△ S = + 285 J K^{- 1} m o l^{- 1}$ . Calculate free energy change $△ G$ at $25^{0}$ C and product whether the reaction is spontaneous or not.

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Solution

Given, $△ H = 177 k J / m o l, △ S = 285 J K^{- 1} m o l^{- 1}, T = 298 K$
Using the relation, $△ G = △ H - T △ S$
$△ G = (177 \times 10^{3} J / m o l) - (298 K) (285 J K^{- 1} m o l^{- 1})$
$△ G = 92070 J / m o l e$
$△ G = 92.07 k J / m o l e$
$△ G > 0$
Since $△ G$ is positive, Hence reaction is non spontaneous.

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For the reaction NH4Cl(S)→NH3(g)+HCl(g) at 250C, enthalpy change △=+177kJmol−1 and entropy change △S=+285JK−1mol−1. Calculate free energy change △G at 250C and product whether the reaction is spontaneous or not.

Given, △H=177 kJ/mol, △S=285 JK−1mol−1, T=298 KUsing the relation, △G=△H−T△S△G=(177×103 J/mol)−(298 K)(285 JK−1mol−1)△G=92070 J/mole△G=92.07 kJ/mole△G>0Since △G is positive, Hence reaction is non spontaneous.

For the reaction $N H_{4} C l (S) \to N H_{3} (g) + H C l (g)$ at $25^{0}$ C, enthalpy change $△ = + 177 k J m o l^{- 1}$ and entropy change $△ S = + 285 J K^{- 1} m o l^{- 1}$ . Calculate free energy change $△ G$ at $25^{0}$ C and product whether the reaction is spontaneous or not.