Question

The equilibrium constant for the reaction A(g)❝B(g) is 10. A reaction mixture initially contains [A] = 1.1 M and [B] = 0.0 M. Which statement is true at equilibrium? The equilibrium constant for the reaction is 10. A reaction mixture initially contains = 1.1 and = 0.0 . Which statement is true at equilibrium?a. The reaction mixture will contain [A] = 1.0 M and [B] = 0.1 M.b. The reaction mixture will contain equal concentrations of A and B.c. The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M.

Answer 1

Answer:

The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M

Explanation:

The reaction is:

• A(g) → B(g)

And the equilibium constant can be expressed in terms of [A] and [B]:

• Keq = 10 = [B]/[A]

If the initial conditions are that [A] = 1.1 M and [B] = 0.0 M, the conditions at equilibrium will be:

• [A] = 1.1M - x

• [B] = 0.0M + x

Now we rewrite Keq:

• Keq = 10 = [0.0 +x] / [1.1 -x]

Finally we solve for x:

• 10 = [0.0 +x] / [1.1 -x]

• 10 * (1.1-x) = x

• 11 - 10x = x

• 11 = 11x

• x = 1

So at equilibrium, the concentrations are:

• [A] = 1.1 M - 1 = 0.1 M
• [B] = 0.0 M + 1 = 1.0 M

Answer 2

The statement that is true at equilibrium is:

a. The reaction mixture will contain [A] = 0.1 M and [B] = 1.0 M

Chemical reaction:

A(g) ⇄ B(g)

Given:

$K_{eq} = 10 =\frac{[B]}{[A]}$

                           [A]                [B]

Initially:               1.1M - x      0.0M + x

Thus, equilibrium constant can be rewritten as:

$K_{eq} = 10 =\frac{ [0.0 +x] }{[1.1 -x]}$

On solving for x:

$10 = \frac{[0.0 +x] }{ [1.1 -x]}\\\\10 * (1.1-x) = x\\\\11 - 10x = x\\\\11 = 11x\\\\x = 1$

So at equilibrium, the concentrations are:

[A] = 1.1 M - 1 = 0.1 M

[B] = 0.0 M + 1 = 1.0 M

Thus, the correct option is a.

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