Answer:
The value of the equilibrium constant KC is 1.244
Explanation:
A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C
At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc
Step 1: The balanced equation
A+2B<->C
Step 2: The initial concentrations
[A] = 0.550 M
[B]= 1.40 M
[C] = 0.600 M
Step 3: The concentraions at equilibrium
[A] = 0.550 -X = 0.430 M
[B]= 1.40 -2X M
[C] = 0.600 + X = 0.720 M
X = 0.120 M
[A] = 0.550 - 0.120 = 0.430 M
[B]= 1.40 -2*0.120 = 1.16 M
[C] = 0.600 + 0.120 = 0.720 M
Step 4: Calculate Kc
Kc = [C] / [A][B]²
Kc = 0.720 / (0.430*1.16²)
Kc = 1.244
The value of the equilibrium constant KC is 1.244
