In the reaction A(s)+B(g)+ heat leftharpoons 2C(s)+2D(g) at equilibrium, pressure of B is doubled to re-establish the equilibrium. The factor by which concentration of D is changed, is:

Question

In the reaction A(s)+B(g)+ heat leftharpoons 2C(s)+2D(g) at equilibrium, pressure of B is doubled to re-establish the equilibrium. The factor by which concentration of D is changed, is: (A)  √2  (B)  2  (C)  3  (D)  2√2

Tardigrade Admin · Accepted Answer

For reaction, Kp=([D]2/[B]) Hence, for this reaction, ([D1]2/[B1])=([D2]2/[2 B1]) or [D2]=√2[D1]

Q. In the reaction $A (s) + B (g) +$ heat $⇌ 2 C (s) + 2 D (g)$ at equilibrium, pressure of $B$ is doubled to re-establish the equilibrium. The factor by which concentration of $D$ is changed, is:

$2$

$2$

$3$

$22$

For reaction, $K_{p} = \frac{[ D ] ^{2}}{[ B ]}$
Hence, for this reaction, $\frac{[ D _{1} ] ^{2}}{[ B _{1} ]} = \frac{[ D _{2} ] ^{2}}{[ 2 B _{1} ]}$
or $[D_{2}] = 2 [D_{1}]$

Q. In the reaction A(s)+B(g)+ heat ⇌2C(s)+2D(g) at equilibrium, pressure of B is doubled to re-establish the equilibrium. The factor by which concentration of D is changed, is:

2​

2

3

22​

For reaction, Kp​=[B][D]2​ Hence, for this reaction, [B1​][D1​]2​=[2B1​][D2​]2​ or [D2​]=2​[D1​]

Q. In the reaction $A (s) + B (g) +$ heat $⇌ 2 C (s) + 2 D (g)$ at equilibrium, pressure of $B$ is doubled to re-establish the equilibrium. The factor by which concentration of $D$ is changed, is:

$2$

$2$

$3$

$22$

For reaction, $K_{p} = \frac{[ D ] ^{2}}{[ B ]}$
Hence, for this reaction, $\frac{[ D _{1} ] ^{2}}{[ B _{1} ]} = \frac{[ D _{2} ] ^{2}}{[ 2 B _{1} ]}$
or $[D_{2}] = 2 [D_{1}]$