Question

Given the chemical reaction below,
A(aq) + B(aq) ⇌ C(aq) + D(l)
When 2.00 mol of A was mixed with 3.00 mol of B in a 1.00 L flask, 0.50 mol of C was formed at room temperature. What is the value of the equilibrium constant, Kc?

A.

0.083

B.

12

C.

8

D.

0.13

Answer 1

Given chemical equation.

A(aq) + B(aq) ⇌ C(aq) + D(l)

Expression for Kc will be given as

Kc = [C] / [A][B] ....... (1) D being Liquid will not appear in expression for Kc.

Now, Initially, [A] = 2 mole / 1L = 2 M & [B] = 3 mole/1L = 3M

After reaction, [C] = 0.5 mole/1L = 0.5M

Hence At equilibrium,

[C] = 0.5 M

[A] = 2 M - 0.5 M = 1.5 M

[B] = 3 M - 0.5 M = 2.5 M

Using these concentrations in eq.(1) we get,

Kc = (0.5) / (1.5*2.5)

Kc = 0.133

Answer option: (D) 0.13

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