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2A (g) + Y (g) <-- --> 3C (g) + D (g)
Based on the initial conditions shown below, determine the value of the equilibrium constant if the concentration of product C at equilibrium was measured to be 0.456 M.
Initial Conditions: No reactants present, [C] = 0.651 M, [D] = 0.754 M.
Equilibrium Conditions: [A] = ?, [Y] = ?, [C] = 0.456 M, [D] = ?.

A. K = 59.5
B. K = 37.2
C. K = 0.0269
D. K = 0.0168

Sagot :

Sebassandinidn

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!

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