Question

2A (g) + Y (g) <-- --> 3C (g) + D (g)

Answer 1

Answer:

A. K = 59.5

Explanation:

Hello there!

In this case, since this reaction seems to start moving leftwards due to the fact that neither A nor Y are present at equilibrium, we should rewrite the equation:

3C (g) + D (g) <-- --> 2A (g) + Y (g)

Thus, the equilibrium expression is:

$K^{left}=\frac{[A]^2[Y]}{[C]^3[D]}$

Next, according to an ICE table for this reaction, we find that:

$[A]=2x$

$[Y]=x$

$[C]=0.651M-3x$

$[D]=0.754M-x$

Whereas x is calculated by knowing that the [C] at equilibrium is 0.456M; thus:

$x=\frac{0.651-0.456}{3} =0.065M$

Next, we compute the rest of the concentrations:

$[A]=2(0.065M)=0.13M$

$[Y]=0.065M$

$[D]=0.754M-0.065M=0.689M$

Thus, the equilibrium constant for the leftwards reaction is:

$K^{left}=\frac{(0.13M)^2(0.065M)}{(0.456M)^3(0.689M)}=0.0168$

Nonetheless, we need the equilibrium reaction for the rightwards reaction; thus, we take the inverse to get:

$K^{right}=\frac{1}{0.0168}=59.5$

Therefore, the answer would be A. K = 59.5.

Regards!