Equilibrium Constant Calculation (Jee Mains)

Second Law & Gibbs Energy

Medium

Equilibrium Constant Calculation

Use the following data :

Substance	$\frac{{{\Delta _f}{H^ \ominus }(500K)}}{{kJ\,mo{l^{-1}}}}$	$\frac{{{S^ \ominus }(500K)}}{{J\,{K^{-1}}\,mo{l^{-1}}}}$
AB(g)	32	222
A₂(g)	6	146
B₂(g)	X	280

One mole each of A₂(g) and B₂(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500K.
A₂(g) + B₂(g) ⇌ 2AB(g)
The value of x (in kJ mol^–1) is ……….. (Nearest integer)
(Given: log K=2.2 R=8.3 JK^–1 mol^–1)

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Second Law & Gibbs Energy

Medium

Equilibrium Constant Calculation

Use the following data :

Substance	$\frac{{{\Delta _f}{H^ \ominus }(500K)}}{{kJ\,mo{l^{-1}}}}$	$\frac{{{S^ \ominus }(500K)}}{{J\,{K^{-1}}\,mo{l^{-1}}}}$
AB(g)	32	222
A₂(g)	6	146
B₂(g)	X	280

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