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}}} $
