Question

If the initial concentrations of A and B are 0M each, and the initial concentrations of C and D are 1.6M each, and Kc=4.7, what is the equilibrium concentration of A?

Answer 1

Answer:

$[A]_{eq}=0.095M$

Explanation:

Hello,

In this case, for the reaction:

$A+B\rightleftharpoons C+D$

It is clear that the equilibrium expression is:

$Kc=\frac{[C][D]}{[A][B]}$

Nevertheless, since the initial concentrations of both A and B are 0 M and C and D 1.6 M, we should invert the reaction:

$C+D\rightleftharpoons A+B$

Thereby the equilibrium expression is also inverted:

$Kc=\frac{[A][B]}{[C][D]}$

Which can be written in terms of the reaction extent and initial concentrations as shown below:

$\frac{1}{Kc} =\frac{x*x}{([C]_0-x)([D]_0-x)}\\\\\frac{1}{0.47}= \frac{x*x}{(1.6-x)(1.6-x)}$

Hence, solving for $x$ we obtain:

$x=0.095M$

In such a way, the equilibrium concentration of A is:

$[A]_{eq}=x=0.095M$

Best regards.

Answer 2

The equilibrium concentration of A will be "0.095 M". To understand the calculation, check below.

Equilibrium

According to the question,

Initial concentration for A and B = 0 M

For C and D = 1.6 M

Kc = 4.7

Here the reaction:

→ A + B $\rightleftharpoons$ C + D

We know the expression,

Kc = $\frac{[C][D]}{[A][B]}$

By inverting the reaction,

→ C + D $\rightleftharpoons$ A + B

Kc = $\frac{[A][B]}{[C][D]}$

By substituting the values,

$\frac{1}{Kc}$ = $\frac{x\times x}{([C]_0 -x)([D]_0 -x)}$

    = $\frac{x\times x}{(1.6 - x)(1.6 -x)}$

 x = 0.095 M ($[A]_{eq}$)

Thus the above approach is correct.

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