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The equilibrium reaction that doesn't have equal values for KcK_c and KpK_p is:

A

2NO(g)N2(g)+O2(g)2NO(g) \rightleftharpoons N_2(g) + O_2(g)

B

SO2(g)+NO2(g)SO3(g)+NO(g)SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)

C

H2(g)+I2(g)2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

D

2C(s)+O2(g)2CO2(g)2C(s) + O_2(g) \rightleftharpoons 2CO_2(g)

Step-by-Step Solution

The relationship between KpK_p and KcK_c is given by Kp=Kc(RT)ΔnK_p = K_c(RT)^{\Delta n}, where Δn\Delta n is the difference between the number of moles of gaseous products and gaseous reactants (Δn=npnr\Delta n = n_p - n_r). For KpK_p to be equal to KcK_c, Δn\Delta n must be zero. Let's calculate Δn\Delta n for the given reactions:

  1. 2NO(g)N2(g)+O2(g)2NO(g) \rightleftharpoons N_2(g) + O_2(g); Δn=(1+1)2=0\Delta n = (1 + 1) - 2 = 0
  2. SO2(g)+NO2(g)SO3(g)+NO(g)SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g); Δn=(1+1)(1+1)=0\Delta n = (1 + 1) - (1 + 1) = 0
  3. H2(g)+I2(g)2HI(g)H_2(g) + I_2(g) \rightleftharpoons 2HI(g); Δn=2(1+1)=0\Delta n = 2 - (1 + 1) = 0
  4. 2C(s)+O2(g)2CO2(g)2C(s) + O_2(g) \rightleftharpoons 2CO_2(g); Δn=21=1\Delta n = 2 - 1 = 1 (Carbon is in the solid state, so it is not considered in calculating Δn\Delta n). Since Δn0\Delta n \neq 0 for reaction 4, its KpK_p and KcK_c values are not equal.
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