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:
[tex]K^{left}=\frac{[A]^2[Y]}{[C]^3[D]}[/tex]
Next, according to an ICE table for this reaction, we find that:
[tex][A]=2x[/tex]
[tex][Y]=x[/tex]
[tex][C]=0.651M-3x[/tex]
[tex][D]=0.754M-x[/tex]
Whereas x is calculated by knowing that the [C] at equilibrium is 0.456M; thus:
[tex]x=\frac{0.651-0.456}{3} =0.065M[/tex]
Next, we compute the rest of the concentrations:
[tex][A]=2(0.065M)=0.13M[/tex]
[tex][Y]=0.065M[/tex]
[tex][D]=0.754M-0.065M=0.689M[/tex]
Thus, the equilibrium constant for the leftwards reaction is:
[tex]K^{left}=\frac{(0.13M)^2(0.065M)}{(0.456M)^3(0.689M)}=0.0168[/tex]
Nonetheless, we need the equilibrium reaction for the rightwards reaction; thus, we take the inverse to get:
[tex]K^{right}=\frac{1}{0.0168}=59.5[/tex]
Therefore, the answer would be A. K = 59.5.
Regards!
