At 700 K hydrogen and bromine react to form hydrogen bromide- The value of equilibrium constant for this reaction is 5- 10 8- Calculate the amount of H2-Br2 and HBr at equilibrium if a mixture of 0-6 mol of H2 and 0-2 mol of bromine is heated to 700 K-

H _ 2 (g)+ Br _ 2 (g)⇌ 2HBr Initial 0.6 #160; 0.2 [ HBr ] #160;^ 2 [ H _ 2 ] [ Br _ 2 ] #160; =K or 4 x ^ 2 (0.6 x)(0.2 x) ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

At 700 K hy...

Question

At $700 K$ hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is $5 \times 10^{8}$ . Calculate the amount of $H_{2}, {B r}_{2}$ and $H B r$ at equilibrium if a mixture of $0.6 m o l$ of $H_{2}$ and $0.2 m o l$ of bromine is heated to $700 K$ .

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Solution

$H_{2} (g) + {B r}_{2} (g) ⇌ 2 H B r$
Initial $0.6$ $0.2$
$\frac{{[H B r]}^{2}}{[H_{2}] [{B r}_{2}]} = K$
or $\frac{4 x^{2}}{(0.6 - x) (0.2 - x)} = 5 \times 10^{8}$
$x = 0.6$ or $0.2$
$2 H B r ⇌ H_{2} (g) + {B r}_{2} (g)$
At equi, $(0.4 - 2 x)$ $(0.4 + x)$ $x$
$x = 2 \times 10^{- 10}$
$[{B r}_{2}] = 2 \times 10^{- 10} m o l$ ;
$[H_{2}] = 0.4 m o l$
$[H B r] = 0.4 m o l$

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At 700 K hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is 5×108. Calculate the amount of H2,Br2 and HBr at equilibrium if a mixture of 0.6 mol of H2 and 0.2 mol of bromine is heated to 700 K.

H2(g)+Br2(g)⇌2HBrInitial 0.6 0.2[HBr]2[H2][Br2]=Kor 4x2(0.6−x)(0.2−x)=5×108x=0.6 or 0.2 2HBr⇌H2(g)+Br2(g)At equi,(0.4−2x) (0.4+x) xx=2×10−10[Br2]=2×10−10mol;[H2]=0.4mol[HBr]=0.4 mol

At $700 K$ hydrogen and bromine react to form hydrogen bromide. The value of equilibrium constant for this reaction is $5 \times 10^{8}$ . Calculate the amount of $H_{2}, {B r}_{2}$ and $H B r$ at equilibrium if a mixture of $0.6 m o l$ of $H_{2}$ and $0.2 m o l$ of bromine is heated to $700 K$ .