If a mixture 0-4 mole H2 and 0-2 mole Br2 is heated at 700 K at equilibrium- the value of equilibrium constant is 0-25 - 1010 then find out the ratio of concentrations of Br2 and HBr Report your answer as Br2HBr- 1011

Nbsp; amp; H_2 amp; + amp; Br_2 amp; ⇌ amp; 2HBr nbsp; t=0 amp; 0.4 amp; nbsp; amp; 0.2 amp; nbsp; amp; nbsp; t=t_eq a ...

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Standard XIII

Chemistry

Effective Equilibrium Constant

If a mixture ...

Question

If a mixture 0.4 mole $H_{2}$ and 0.2 mole $B r_{2}$ is heated at $700 K$ at equilibrium, the value of equilibrium constant is $0.25 \times 10^{10}$ then find out the ratio of concentrations of $(B r_{2})$ and $(H B r)$ (Report your answer as $\frac{B r_{2}}{H B r} \times 10^{11}$ )

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Solution

The correct option is B 80
$\begin{matrix} H_{2} & + & B r_{2} & ⇌ & 2 H B r t = 0 & 0.4 & 0.2 & - t = t_{eq} & 0.2 & y & 0.4 = negligible & = y \end{matrix}$
$∵ \frac{1}{4} \times 10^{10} = \frac{0.4 \times 0.4}{0.2 \times y} \Rightarrow y = 3.2 \times 10^{- 10} \Rightarrow \frac{B r_{2}}{H B r} \times 10^{11} = \frac{3.2}{0.2} \times 10^{- 10} \times 10^{11} = 80$

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Effective Equilibrium Constant

Standard XIII Chemistry

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If a mixture 0.4 mole H2 and 0.2 mole Br2 is heated at 700 K at equilibrium, the value of equilibrium constant is 0.25×1010 then find out the ratio of concentrations of (Br2) and (HBr) (Report your answer as Br2HBr×1011)

The correct option is B 80H2+Br2⇌2HBrt=00.40.2−t=teq0.2y0.4=negligible=y ∵14×1010=0.4×0.40.2×y⇒y=3.2×10−10⇒Br2HBr×1011=3.20.2×10−10×1011=80

If a mixture 0.4 mole $H_{2}$ and 0.2 mole $B r_{2}$ is heated at $700 K$ at equilibrium, the value of equilibrium constant is $0.25 \times 10^{10}$ then find out the ratio of concentrations of $(B r_{2})$ and $(H B r)$ (Report your answer as $\frac{B r_{2}}{H B r} \times 10^{11}$ )