Given the reaction between 2 gases represented by A2 and B2 to give the compound ABg-A2g-B2g- 2ABgAt equilibrium- the concentration of-A2-3-0- 10-3 MB2-4-2- 10-3 MAB-2-8- 10-3 MIf the reaction takes place in a sealed vessel at 527 -C- then the value of Kc will be-

The equilibrium reaction is as given below. A_2(g)+B_2(g)⇌ 2AB(g) The expression for the equilibrium constant is as given below: K_c=[AB]^2[A_2][B_2] Substitute ...

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Byju's Answer

Standard XII

Chemistry

Effect on Kc

Given the rea...

Question

Given the reaction between $2$ gases represented by $A_{2}$ and $B_{2}$ to give the compound $A B (g)$ .
$A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)$
At equilibrium, the concentration of:
$A_{2} = 3.0 \times 10^{- 3} M$
$B_{2} = 4.2 \times 10^{- 3} M$
$A B = 2.8 \times 10^{- 3} M$
If the reaction takes place in a sealed vessel at $527^{\circ} C$ , then the value of $K_{c}$ will be:

2.0

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1.9

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0.62

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4.5

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Solution

The correct option is C $0.62$
The equilibrium reaction is as given below.
$A_{2} + B_{2} ⇌ 2 A B$
The expression for the equilibrium constant is as given below:
$K_{c} = \frac{[A B]^{2}}{[A_{2}] [B_{2}]}$
Substitute the values in the above expression
$K_{c} = \frac{2.8 \times 2.8 \times 10^{- 6}}{3 \times 10^{- 3} \times 4.2 \times 10^{- 3}} = 0.62$
Hence, the value of the equilibrium constant is $0.62$ .

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Given the reaction between 2 gases represented by A2 and B2 to give the compound AB(g). A2(g)+B2(g)⇌2AB(g) At equilibrium, the concentration of: A2=3.0×10−3 M B2=4.2×10−3 M AB=2.8×10−3 M If the reaction takes place in a sealed vessel at 527 ∘C, then the value of Kc will be: