If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

Question

PilotLPTM

28 days ago

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If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

M SO2, 0.500 M NO2, 0.00500 M SO3, and 0.00500 M NO?

Chemistry

1 answer:

VMariaS [2.7K] · Answer 1 · 2026-03-09T08:52:56+03:00

Answer: The concentrations at equilibrium for sulfur dioxide, nitrogen dioxide, sulfur trioxide, and nitrogen monoxide are 0.196 M, 0.196 M, 0.309 M, and 0.309 M, respectively.

Explanation:

Initial concentrations provided are:

Sulfur dioxide = 0.500 M

Nitrogen dioxide = 0.500 M

Sulfur trioxide = 0.00500 M

Nitrogen monoxide = 0.00500 M

The following reaction occurs:

$SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial conditions: 0.500 0.500 0.005 0.005

At equilibrium: 0.500-x 0.500-x 0.005+x 0.005+x

The equilibrium constant expression derived from the reaction is:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

Given:

$K_c=2.50$

Inserting values into the equation leads to:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Ignoring x = 1.37, since the change must not exceed the initial concentration

Therefore, the concentration of sulfur dioxide at equilibrium = $(0.500-x)=(0.500-0.304)=0.196M$

Nitrogen dioxide equilibrium concentration = $(0.500-x)=(0.500-0.304)=0.196M$

Sulfur trioxide equilibrium concentration = $(0.00500+x)=(0.00500+0.304)=0.309M$

Nitrogen monoxide equilibrium concentration = $(0.00500+x)=(0.00500+0.304)=0.309M$

Thus, the equilibrium concentrations of sulfur dioxide, nitrogen dioxide, sulfur trioxide, and nitrogen monoxide are 0.196 M, 0.196 M, 0.309 M, and 0.309 M, correspondingly.