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Question

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.

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Solution

H2(g)+Br2(g)2HBr
Initial 0.6 0.2
[HBr]2[H2][Br2]=K
or 4x2(0.6x)(0.2x)=5×108
x=0.6 or 0.2
2HBrH2(g)+Br2(g)
At equi,(0.42x) (0.4+x) x
x=2×1010
[Br2]=2×1010mol;
[H2]=0.4mol
[HBr]=0.4 mol

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