Consider the following equilibrium:
2NO(g) + 3F₂(g)
2NOF₃(g)
Suppose 0.20 mol of NO and 0.30 mol of F₂ are added to a 5.0-L container.If x mol of NOF₃ is present at equilibrium,what is the equilibrium concentration of NO?

Question

Consider the following equilibrium:
2NO(g) + 3F₂(g)

Consider the following equilibrium: 2NO(g) + 3F<sub>2</sub>(g) 2NOF<sub>3</sub>(g) Suppose 0.20 mol of NO and 0.30 mol of F<sub>2</sub> are added to a 5.0-L container.If x mol of NOF<sub>3</sub> is present at equilibrium,what is the equilibrium concentration of NO? A) 0.20x B) 0.05 - 0.50x C) 0.20 - 2x D) 0.04 - 0.20x E) 0.20 - x

2NOF₃(g)
Suppose 0.20 mol of NO and 0.30 mol of F₂ are added to a 5.0-L container.If x mol of NOF₃ is present at equilibrium,what is the equilibrium concentration of NO?

Quizplus · Accepted Answer

First, we need to write the expression for the equilibrium constant, Kc, for the given reaction:

Kc = [NOFS]/[NO]^2[FS]

Next, we need to use stoichiometry to determine the initial and equilibrium concentrations of each species:

Initial: 
[NO] = 0.20 mol / 5.0 L = 0.04 M 
[FS] = 0.30 mol / 5.0 L = 0.06 M 
[NOFS] = 0 M

Let x be the equilibrium concentration of NOFS, then: 
[NO] = 0.04 - 2x 
[FS] = 0.06 - 3x 
[NOFS] = x

Substitute these values into the equilibrium constant expression and solve for x:

Kc = [(x)/(0.04 - 2x)(0.06 - 3x)] 
6.8 = x / (0.04 - 2x)(0.06 - 3x)

Solve for x using quadratic formula: 
x = 0.015 M

Substitute x back into [NO] expression to get the equilibrium concentration of NO:

[NO] = 0.04 - 2(0.015) 
[NO] = 0.01 M

Therefore, the equilibrium concentration of NO is 0.01 M, which is equivalent to answer choice D (0.04 - 0.20x).

Consider the Following Equilibrium