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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The equilibri...

Question

The equilibrium constant $K_{p}$ (in atm) for the reaction is $9$ at $7 a t m$ and $300 K$
$A_{2} (g) ⇌ B_{2} (g) + C_{2} (g)$
Calculate the average molar mass (in gm/mol) of an equilibrium mixture:
Given Molar of $A_{2}, B_{2}$ and $C_{2}$ are $70, 49$ & $21$ gm/ mol respectively.

50

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Solution

The correct option is A $50$
$K p = K c (R T)^{Δ n}$
since $Δ n = 1$ , then $K c (R T)$
$9 = K c \times 0.0821 \times 300$
$K c = 0.3654$
$K c = \frac{[A] [B]}{[C]} = \frac{X^{2}}{1} = 0.3654$
$x = 0.6044$
average molecular weight $= 70 \times 0.455 + 49 \times 0.27 + 21 \times 0.27 = 50$
$∴ (A)$ is correct option

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The equilibrium constant Kp (in atm) for the reaction is 9 at 7 atm and 300KA2(g)⇌B2(g)+C2(g)Calculate the average molar mass (in gm/mol) of an equilibrium mixture:Given Molar of A2,B2 and C2 are 70,49 & 21 gm/ mol respectively.

The correct option is A 50Kp=Kc(RT)Δnsince Δn=1, then Kc(RT)9=Kc×0.0821×300Kc=0.3654Kc=[A][B][C]=X21=0.3654x=0.6044average molecular weight =70×0.455+49×0.27+21×0.27=50∴ (A) is correct option

The equilibrium constant $K_{p}$ (in atm) for the reaction is $9$ at $7 a t m$ and $300 K$
$A_{2} (g) ⇌ B_{2} (g) + C_{2} (g)$
Calculate the average molar mass (in gm/mol) of an equilibrium mixture:
Given Molar of $A_{2}, B_{2}$ and $C_{2}$ are $70, 49$ & $21$ gm/ mol respectively.