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JEE Main 201912 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
12 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Equilibrium

JEE Main 2019 · 12 April · Evening Shift · Chemistry · Chemical Equilibrium

In which one of the following equilibria, K _ p ≠ K _ c ?

In which one of the following equilibria, Kp≠Kc?{K}_{p} \ne {K}_{c} ?

Options

  1. A2HI(g)⇌H2(g)+I2(g)2HI\left(g\right) ⇌ {H}_{2}\left(g\right) + {I}_{2}\left(g\right)
  2. B2C(s)+O2(g)⇌2CO(g)2C\left(s\right) + {O}_{2}\left(g\right) ⇌ 2CO\left(g\right)
  3. C2NO(g)⇌N2(g)+O2(g)2NO\left(g\right) ⇌ {N}_{2}\left(g\right) + {O}_{2}\left(g\right)
  4. DNO2(g)+SO2(g)⇌NO(g)+SO3(g)N{O}_{2}\left(g\right) + S{O}_{2}\left(g\right) ⇌ NO\left(g\right) + S{O}_{3}\left(g\right)

Correct answer

B. 2C(s)+O2(g)⇌2CO(g)2C\left(s\right) + {O}_{2}\left(g\right) ⇌ 2CO\left(g\right)

Step-by-step solution

kp=kc(RT)Δng{k}_{p} = {k}_{c}{\left(RT\right)}^{\Delta{n}_{g}}
(i)2HI(g)⇌H2(g)+I2(g)( i ) 2HI\left(g\right) ⇌ {H}_{2}\left(g\right) + {I}_{2}\left(g\right)
Δng=1+1−2=0\Delta{n}_{g} = 1 + 1 - 2 = 0
Sokp=kcSo{k}_{p} = {k}_{c}
(ii) 2NO(g)⇌N2(g)+O2(g)2NO\left(g\right) ⇌ {N}_{2}\left(g\right) + {O}_{2}\left(g\right)
Δng=2−2=0\Delta{n}_{g} = 2 - 2 = 0
Sokp=kcSo{k}_{p} = {k}_{c}
(iii) NO2(g)+SO2(g)⇌NO(g)+SO3(g){NO}_{2}\left(g\right) + {SO}_{2}\left(g\right) ⇌ NO\left(g\right) + {SO}_{3}\left(g\right)
Δng=2−2=0\Delta{n}_{g} = 2 - 2 = 0
Sokp=kcSo{k}_{p} = {k}_{c}
(iv) 2(s)+O2(g)⇌2CO(g)2\left(s\right) + {O}_{2}\left(g\right) ⇌ 2CO\left(g\right)
Δng=2−1=1\Delta{n}_{g} = 2 - 1 = 1
kp=kc(RT)\Deltan⇒kp=kc(RT){k}_{p} = {k}_{c}{\left(\text{RT}\right)}^{\Deltan} ⇒ {k}_{p} = {k}_{c}\left(\text{RT}\right)
kp>kc{k}_{p} > {k}_{c}
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