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InterviewSolution
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When 3.00 mole of A and 1.00 mole of B are mixed ina 1.00 litre vessel, the following reaction takes placeA(g) + B(g) 20(g)The equilibrium mixture contains 0.5 mole of C. Whatthe value of equilibrium constant for the reaction?(2) 6(4) 3(1) 0.12 |
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Answer» In the given question, volume=1L, so initial concentration of A=3mol/1L=3MSimilarly for B=1MNow,A(g) + B(g)= 2C(g)3mol/L.....1mol/L..0.......(initially)3-x..........1-x …......2x...at equilibriumBut given that, mole of C=0.5 at equilibrium, Hence its equilibrium concentration=0.5/1L=0.5M=2x(from our equation)So x=0.25Mwriting for the equilibrium constantK=[C]2/[A]*[B]=(2x)2/(3-x)*(1-x)Put here the value of x and calculate KK=0.12 Note: we should always put the concentratin with their unit |
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