In the reaction,
$A (s) + B (g) + H e a t ⇌ 2 C (s) + 2 D (g)$
equilibrium is established. The pressure of D is increased and found be doubled when equilibrium is restablished. The factor by which B is changed is:

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Solution

The correct option is C 4
Assume at equilibrium, the pressures of B and D are 1 and 2 atm respectively.
$K = (1) (2)^{2} = 4$
After second equilibrium, the pressure of C is doubled to 4.
$K = (P_{B}) (4)^{2} = 4$
$P_{B} = \frac{1}{4}$ .
Thus, the pressure of B is reduced by a factor of 4.

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In the reaction,A(s)+B(g)+Heat⇌2C(s)+2D(g)equilibrium is established. The pressure of D is increased and found be doubled when equilibrium is restablished. The factor by which B is changed is:

The correct option is C 4Assume at equilibrium, the pressures of B and D are 1 and 2 atm respectively.K=(1)(2)2=4After second equilibrium, the pressure of C is doubled to 4.K=(PB)(4)2=4PB=14.Thus, the pressure of B is reduced by a factor of 4.

In the reaction,
$A (s) + B (g) + H e a t ⇌ 2 C (s) + 2 D (g)$
equilibrium is established. The pressure of D is increased and found be doubled when equilibrium is restablished. The factor by which B is changed is:

The correct option is C 4
Assume at equilibrium, the pressures of B and D are 1 and 2 atm respectively.
$K = (1) (2)^{2} = 4$
After second equilibrium, the pressure of C is doubled to 4.
$K = (P_{B}) (4)^{2} = 4$
$P_{B} = \frac{1}{4}$ .
Thus, the pressure of B is reduced by a factor of 4.