Question:

For which of the following reactions, $K_{p}=K_{c} ?$

Updated On: Jun 8, 2024

$N _{2}+3 H _{2} 2 NH _{3}$
$N _{2}+ O _{2} 2 NO$
$PCl _{5} PCl _{3}+ Cl _{2}$
$2 SO _{3} 2 SO _{2}+ O _{2}$

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The Correct Option is B

Solution and Explanation

$K_{p}=K_{c}(R T)^{\Delta n}$
For the reaction: $N_{2}+O_{2} 2 N O$
$\Delta n=2-2=0 $
$\therefore K_{p}=K_{c}(R T)^{0} $
$K_{p}=K_{c}$

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Concepts Used:

Law of Chemical Equilibrium

Law of Chemical Equilibrium states that at a constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants each raised to a power equal to the corresponding stoichiometric coefficients as represented by the balanced chemical equation.

Let us consider a general reversible reaction;

A+B ↔ C+D

After some time, there is a reduction in reactants A and B and an accumulation of the products C and D. As a result, the rate of the forward reaction decreases and that of backward reaction increases.

Eventually, the two reactions occur at the same rate and a state of equilibrium is attained.

By applying the Law of Mass Action;

The rate of forward reaction;

R_f = K_f [A]^a [B]^b

The rate of backward reaction;

R_b = K_b [C]^c [D]^d

Where,

[A], [B], [C] and [D] are the concentrations of A, B, C and D at equilibrium respectively.

a, b, c, and d are the stoichiometric coefficients of A, B, C and D respectively.

Kf and Kb are the rate constants of forward and backward reactions.

However, at equilibrium,

Rate of forward reaction = Rate of backward reaction.

Kc is called the equilibrium constant expressed in terms of molar concentrations.

The above equation is known as the equation of Law of Chemical Equilibrium.