The question is incomplete, here is the complete question:
A(aq) + B(aq) → 2C(aq)
If the value of Kc for the reaction is 434, what is the concentration of C at equilibrium if initial concentrations of A and B are both 0.500 M. (Hint: Everything is squared after you set-up the equilibrium expression with the values given.
<u>Answer:</u> The concentration of C at equilibrium is 0.912 M
<u>Explanation:</u>
We are given:
Initial concentration of A = 0.500 M
Initial concentration of B = 0.500 M
The given equation follows:
![A(aq.)+B(aq.)\rightarrow 2C(aq.)](https://tex.z-dn.net/?f=A%28aq.%29%2BB%28aq.%29%5Crightarrow%202C%28aq.%29)
<u>Initial:</u> 0.5 0.5
<u>At eqllm:</u> 0.5-x 0.5-x 2x
The expression of
for above equation follows:
![K_c=\frac{[C]^2}{[A][B]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BC%5D%5E2%7D%7B%5BA%5D%5BB%5D%7D)
We are given:
![K_c=434](https://tex.z-dn.net/?f=K_c%3D434)
Putting values in above equation, we get:
![434=\frac{(2x)^2}{(0.5-x)\times (0.5-x)}\\\\x=0.456,0.553](https://tex.z-dn.net/?f=434%3D%5Cfrac%7B%282x%29%5E2%7D%7B%280.5-x%29%5Ctimes%20%280.5-x%29%7D%5C%5C%5C%5Cx%3D0.456%2C0.553)
Neglecting the value of x = 0.553 M because the equilibrium concentration of A and B will become negative, which is not possible
So, equilibrium concentration of C = 2x = 2(0.456) = 0.912 M
Hence, the concentration of C at equilibrium is 0.912 M