Ncert Exercises & Solutions

The value of $∆ n$ for the following reaction will be :

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

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.

Step 1:

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)

Step 2:

For solid-state reactants 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