Chemistry: Amixture initially contains a, b, and c in the following concentrations: [a] = 0.350 m , [b] = 0.650 m , and [c] = 0.300 m . the following reaction occurs and equilibrium is established: a+2b⇌c at equilibrium, [a] = 0.220 m and [c] = 0.430 m . calculate the value of the equilibrium constant, kc.

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Amixture initially contains a, b, and c in the

nickname9154

12.03.2022

The value of equilibrium constant, K_c for the given reaction is 12.85.

Explanation:

For the given chemical equation:

$A+2B\rightleftharpoons C$

At t = 0 0.350M 0.650M 0.300M

At $t=t_{eq}$ (0.350 - x) (0.650 - 2x) (0.300 + x)

We are given:

Equilibrium concentration of A = 0.220 M

Forming an equation for concentration of A at equilibrium:

$0.350-x=0.220\\x=0.130$

Thus, the concentration of B at equilibrium becomes = $0.650-(2\times 0.130)=0.390M$

Equilibrium concentration of C = 0.430 M

The expression of K_c for the given chemical equation is:

$K_c=\frac{[C]}{[A][B]^2}$

Putting values in above equation:

$K_c=\frac{0.430}{0.220\times (0.390)^2}\\\\K_c=12.85$

Hence, the value of equilibrium constant, K_c for the given reaction is 12.85.

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