Question

When `2.82` g of solid `NH_(4)Cl` is intoroduced into a 2 L flask at `30^(@)C.` 40 per cent of the solid `NH_(4)Cl` decomposes into two gaseous products, that is `NH_(3)` and HCl. Calculate the Kc. What would happen if more amount of `NH_(4)Cl` is introduced into the flask ?

Answer 1

`NH_(4)Cl_((s)) hArr NH_(3(g))+ HCl_((g))`
2.82 g of `NH_(4)Cl` is subjected to decomposition.
Molecular weight of `NH_(4)Cl=14+4+35.5=53.5`
`implies` No. of moles of `NH_(4)Cl=(2.82)/(53.5)=0.0527`
Given that percent dissociation of `NH_(4)Cl` is 40 percent 0.0527 moles decomposition corresponds to 100 per cent? moles decomposition corresponding to 40 per cent `=(40xx0.0527)/(100)=0.0210`
`NH_(4)Cl hArr NH_(3) + HCl`
`{:("Initial number of moles" ,0.0527,0,0),("Number of moles ",0.0527,,-0.0210),(,0.0210,,0.0210):}`
at equilibrium =0.317
`therefore K_(c)=[NH_(3)][HCl]=(0.0210)/(2)xx(0.0210)/(2)`
`=(0.0210xx0.0210)/(4)=1.1xx10^(-4) "mol"^(-4)//L^(2)`
Addition of a solid or liquid reactant at equilibrium does not affect the equilibrium . So addition of solid `NH_(4)Cl` does not affect the equilibrium .