Suppose $"2.83 g"$ of ammonium chloride solid is placed into a closed, evacuated rigid container, and allowed to decompose at a certain temperature. If 40% of it decomposed into ammonia and hydrogen chloride gas, what is $K_c$ for this reaction?

Md Ariful Islam

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Md Ariful Islam

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The first step in any equilibrium problem like this is to write the complete balanced equation:

$NH_4Cl(s) harr NH_3(g) + HCl(g)$

The form of $K_c$ for this reaction (the subscript c means concentration, and the standard concentration is in units of $"mol/L"$) is:

$K_c = [NH_3][HCl]$

The $NH_4Cl$ does not appear in the equilibrium expression because it is a solid and has a thermodynamic activity of 1.

The starting amount of reactant is $(2.83g)/(59.49 g/(mol))=0.0476 mol$

Because 40% of the reactant decomposed, the equilibrium concentrations are:
$[NH_3]_e = [HCl]_e = (0.0476mol times 0.40)/(2L)=0.00951 M$

The value of $K_c$ is therefore $K_c=(0.00951)^2=9.05 times 10^-5$

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