The value of $∆\mathrm{n}$ for the following reaction will be :

${\mathrm{NH}}_{4}\mathrm{Cl}\left(\mathrm{s}\right)⇌{\mathrm{NH}}_{3}\left(\mathrm{g}\right)+\mathrm{HCl}\left(\mathrm{g}\right)$

1. 1

2. 0.5

3. 1.5

4. 2

Hint: $∆{\mathrm{n}}_{\mathrm{g}}$ =  number of gaseous molecules of products - number of gaseous molecules of reactants.
Step 1:
The relationship between ${\mathrm{K}}_{\mathrm{c}}$ and ${\mathrm{K}}_{\mathrm{p}}$ is
${\mathrm{K}}_{\mathrm{p}}={\mathrm{K}}_{\mathrm{c}}{\left(\mathrm{RT}\right)}^{∆\mathrm{n}}$
where, $∆\mathrm{n}=$(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,
${\mathrm{NH}}_{4}\mathrm{Cl}\left(\mathrm{s}\right)⇌{\mathrm{NH}}_{3}\left(\mathrm{g}\right)+\mathrm{HI}\left(\mathrm{g}\right)\phantom{\rule{0ex}{0ex}}∆\mathrm{n}=2-0=2$