Question

Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:

2HNO3 (aq) + NO (g) → 3NO2 (g) + H2O (l)

At a certain temperature, a chemist finds that a 9.5L reaction vessel containing a mixture of nitric acid, nitrogen monoxide, nitrogen dioxide, and water at equilibrium has the following composition:

Compound Amount
HNO3 15.5g
NO 16.6g
NO2 22.5g
H2O 189.0g

Required:
Calculate the value of the equilibrium constant Kc for this reaction.

Answer 1

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Step-by-step explanation:

Equilibrium concentration of
`HNO_3` =
`(15.5g)/(63g/mol* 9.5L)=0.026M`

Equilibrium concentration of
`NO` =
`(16.6g)/(30g/mol* 9.5L)=0.058M`

Equilibrium concentration of
`NO_2` =
`(22.5g)/(46g/mol* 9.5L)=0.051M`

Equilibrium concentration of
`H_2O` =
`(189.0g)/(18g/mol* 9.5L)=1.10M`

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
`K_c`

For the given chemical reaction:

`2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)`

The expression for
`K_c` is written as:

`K_c=([NO_2]^3* [H_2O]^1)/([HNO_3]^2* [NO]^1)`

`K_c=((0.051)^3* (1.10)^1)/((0.026)^2* (0.058)^1)`

`K_c=3.72`

Thus the value of the equilibrium constant Kc for this reaction is 3.72