Ncert Exercises & Solutions

We know that the relationship between $K_{c}$ and $K_{p}$ is

$K_{p} = K_{c} {(RT)}^{∆ n}$

What would be the value of $∆ n$ for the reaction?

${NH}_{4} Cl (s) ⇌ {NH}_{3} (g) + HI (g)$

(1) 1

(2) 0.5

(3) 1.5

(4) 2

Hint:

∆ n_{g}

= number of gaseous molecules of products - number of gaseous molecules of reactants.

The relationship between

K_{c}

and

K_{p}

K_{p} = K_{c} {(RT)}^{∆ n}

where,

∆ n =

(number of moles of gaseous products)-(number of moles of gaseous reactants)

For solide state reactant or product number of moles is constant and we did not include it.

For the reaction,

{NH}_{4} Cl (s) ⇌ {NH}_{3} (g) + HI (g) ∆ n = 2 - 0 = 2