Equilibrium Question 571

Question: For the reversible reaction, $ {N_{2(g)}}+3{H_{2(g)}} $ ⇌ $ 2N{H_{3(g)}} $ at 500°C, the value of $ K_{P} $ is $ 1.44\times {10^{-5}} $ when partial pressure is measured in atmospheres. The corresponding value of $ K_{c} $ with concentration in mole litre-1, is [IIT Screening 2000; Pb. CET 2004]

Options:

A) $ 1.44\times {10^{-5}} $ / $ {{( 0.082\times 500 )}^{-2}} $

B) $ 1.44\times {10^{-5}} $ / $ {{( 8.314\times 773 )}^{-2}} $

C) $ 1.44\times {10^{-5}} $ / $ {{( 0.082\times 773 )}^{2}} $

D) $ 1.44\times {10^{-5}} $ / $ {{( 0.082\times 773 )}^{-2}} $

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Answer:

Correct Answer: D

Solution:

$ \underset{4}{\mathop{N_2+3H_2}}, $ ⇌ $ \underset{2}{\mathop{2NH_3}}, $

$ \Delta n $ = 2 - 4 = - 2 $ K_{p}=K_{c}{{[RT]}^{\Delta n}} $ ; $ K_{p}=K_{c}{{[RT]}^{-2}} $

$ K_{c}=\frac{K_{p}}{{{[RT]}^{-2}}}=\frac{1.44\times {10^{-5}}}{{{[0.082\times 773]}^{-2}}} $



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