What will be the melting point of KCl if enthalpy change for the reaction is 7.25 J $m o l^{- 1}$ and entropy change is 0.007 J $K^{- 1} m o l^{- 1}$ ?

1835.2 K

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173 K

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1035.7 K

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1285.2 K

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Solution

The correct option is C 1035.7 K
The given reaction is :-
$K C l (s) ⇌ K C l (l)$
Now, at equilibrium, $△ G = 0$
So, from fibb's free energy change equation,
$△ G = △ H - T △ S$
$\Rightarrow △ H = T △ S$
$\Rightarrow T = \frac{△ H}{△ S} = \frac{7.25}{0.007} = 1035.7 k$

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What will be the melting point of KCl if enthalpy change for the reaction is 7.25 J mol−1 and entropy change is 0.007 J K−1 mol−1?

The correct option is C 1035.7 KThe given reaction is :-KCl(s)⇌KCl(l)Now, at equilibrium, △G=0So, from fibb's free energy change equation, △G=△H−T△S⇒△H=T△S⇒T=△H△S=7.250.007=1035.7k

What will be the melting point of KCl if enthalpy change for the reaction is 7.25 J $m o l^{- 1}$ and entropy change is 0.007 J $K^{- 1} m o l^{- 1}$ ?

The correct option is C 1035.7 K
The given reaction is :-
$K C l (s) ⇌ K C l (l)$
Now, at equilibrium, $△ G = 0$
So, from fibb's free energy change equation,
$△ G = △ H - T △ S$
$\Rightarrow △ H = T △ S$
$\Rightarrow T = \frac{△ H}{△ S} = \frac{7.25}{0.007} = 1035.7 k$