Question

A solution is made by mixing 30.0 mL of 0.150 M compound A with 25.0 mL of 0.200 M compound B. At equilibrium, the concentration of C is 0.0454 M. Calculate the equilibrium constant, K, for this reaction.

Answer 1

To calculate the equilibrium constant, K, for the reaction A(aq) + 2B(aq) ⇒ 2C(aq), we can use the concentrations of A, B, and C at equilibrium. The equilibrium constant (K) for this reaction is approximately 5.07 x 10^-3.

Step-by-step explanation:

To calculate the equilibrium constant, K, for the reaction A(aq) + 2B(aq) ⇒ 2C(aq), we need to use the concentrations of A, B, and C at equilibrium. Given that the equilibrium concentration of C is 0.0454 M, and the initial concentrations of A and B are 0.150 M and 0.200 M respectively, we can set up an expression for K.

The mathematical expression for the equilibrium constant, Kc, is given by:

Kc = [C]² / ([A] * [B]²)

Substituting the given equilibrium concentration and initial concentrations into the expression, we get:

Kc = (0.0454)² / (0.150 * (0.200)²) = 5.06666667 x 10^-3

Therefore, the equilibrium constant (K) for this reaction is approximately 5.07 x 10^-3.

Answer 2

Equilibrium Constant K=6.10574

Step-by-step explanation:

Consider the general form of reaction:

aA+bB⇌ cC+dD

Equilibrium Constant of above reaction is:

`K=([C]^c[D]^d)/([A]^a[B]^b)`

Where:

[C] and [D] is the concentration of Products

[A] and [B] is the concentration of reactants

a,b,c,d are number of moles

In our case above reaction will become:

A+B⇌ C

Total Concentration of reactant=0.030 L+0.025 L=0.055 L

Equilibrium Constant will become:

`K=([C]^)/([A][B])`

Now,

`[A]=((0.030L)*(0.150))/(0.03+0.025)=0.0818M`

`[B]=((0.025L)*(0.200))/(0.03+0.025)=0.0909M`

`[C]=0.0454 M`

`K=(0.0454)/((0.0818*0.0909)) \\K=6.10574`

Equilibrium Constant K=6.10574