Suppose $2.83 g$ $"2.83 g"$ 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 $K_{c}$ $K_c$ for this reaction?

Question

Suppose $2.83 g$ $"2.83 g"$ 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 $K_{c}$ $K_c$ for this reaction?

1 Answer

Impact of this question

2364 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-04T23:38:41

The equilibrium constant, $K_{c} = 9.05 \times 10^{- 5}$ $K_c=9.05 times 10^-5$ is calculated from the equilibrium concentrations of gases. If more $N H_{4} C l (s)$ $NH_4Cl(s)$ is added, nothing happens because the gases are already at their equilibrium concentrations.

Explanation:

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

$N H_{4} C l (s) \leftrightarrow N H_{3} (g) + H C l (g)$ $NH_4Cl(s) harr NH_3(g) + HCl(g)$

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

$K_{c} = [N H_{3}] [H C l]$ $K_c = [NH_3][HCl]$

The $N H_{4} C l$ $NH_4Cl$ 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 $\frac{2.83 g}{59.49 \frac{g}{m o l}} = 0.0476 m o l$ $(2.83g)/(59.49 g/(mol))=0.0476 mol$

Because 40% of the reactant decomposed, the equilibrium concentrations are:
${[N H_{3}]}_{e} = {[H C l]}_{e} = \frac{0.0476 m o l \times 0.40}{2 L} = 0.00951 M$ $[NH_3]_e = [HCl]_e = (0.0476mol times 0.40)/(2L)=0.00951 M$

The value of $K_{c}$ $K_c$ is therefore $K_{c} = {(0.00951)}^{2} = 9.05 \times 10^{- 5}$ $K_c=(0.00951)^2=9.05 times 10^-5$

Science

Math

Humanities

... and beyond

Suppose $2.83 g$ $"2.83 g"$ 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 $K_{c}$ $K_c$ for this reaction?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Suppose 2.83 g"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 KcK_c for this reaction?

1 Answer

Explanation:

Related questions

Impact of this question

Suppose $2.83 g$ $"2.83 g"$ 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 $K_{c}$ $K_c$ for this reaction?