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Standard XII

Chemistry

Characteristics of Equilibrium Constant

In a gaseous ...

Question

In a gaseous reaction $A + 2 B ⇌ 2 C + D$ , the initial concentration of $B$ was $1.5$ times that of $A$ . At equilibrium, the concentrations of $A$ and $D$ were equal. The equilibrium constant $K_{c}$ is:

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Solution

Assume initial conentration of $B$ to be $3$ and $A$ to $2$ .

$A + 2 B ⇌ 2 C + D$

Initial $2$ $3$ - -
At eqm $2 - 1 = 1$ $3 - 2 = 1$ $2$ $1$

So, $K_{c} = \frac{2 \times 2 \times 1}{1 \times 1 \times 1} = 4$

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The reaction A + 2B $⇌$ 2C + D is carried out at 298 K using only A and B initially. While mixing, the concentration of B was 1.5 times of that of A. When this system attained equilibrium, it was observed that the (new) concentrations of A and D were same. What is the value of $K_{c}$ (equilibrium constant) at 298 K:___

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In a gaseous reaction A+2B⇌2C+D, the initial concentration of B was 1.5 times that of A. At equilibrium, the concentrations of A and D were equal. The equilibrium constant Kc is:

Assume initial conentration of B to be 3 and A to 2. A+2B⇌2C+DInitial 2 3 - -At eqm 2−1=1 3−2=1 2 1So, Kc=2×2×11×1×1=4

In a gaseous reaction $A + 2 B ⇌ 2 C + D$ , the initial concentration of $B$ was $1.5$ times that of $A$ . At equilibrium, the concentrations of $A$ and $D$ were equal. The equilibrium constant $K_{c}$ is:

Assume initial conentration of $B$ to be $3$ and $A$ to $2$ .

$A + 2 B ⇌ 2 C + D$

Initial $2$ $3$ - -
At eqm $2 - 1 = 1$ $3 - 2 = 1$ $2$ $1$

So, $K_{c} = \frac{2 \times 2 \times 1}{1 \times 1 \times 1} = 4$