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Equilibrium

Question
CBSEENCH11006383

A mixture of straight H subscript 2 comma space straight N subscript 2 space and space NH subscript 3 with molar concentration 3.0 space cross times space 10 to the power of negative 3 end exponent space mol space straight L to the power of negative 1 end exponent comma space 1.0 space cross times space 10 to the power of negative 3 end exponent space mol space straight L to the power of negative 1 end exponent and 2.0 space cross times space 10 cubed space mol space straight L to the power of negative 1 end exponent respectively was prepared at 500 K. At this temperature the value of straight K subscript straight c for reaction straight N subscript 2 left parenthesis straight g right parenthesis space plus space 3 straight H subscript 2 space rightwards harpoon over leftwards harpoon space space space 2 NH subscript 3 left parenthesis straight g right parenthesis comma space space is space 61. Predict whether at this state the concentration of NH subscript 3 will increases or decrease.

Solution
Reaction quotient, Qc for the reaction will be written as
space straight Q subscript straight c space equals space fraction numerator open square brackets NH subscript 3 left parenthesis straight g right parenthesis close square brackets squared over denominator open square brackets straight N subscript 2 left parenthesis straight g right parenthesis close square brackets space open square brackets straight H subscript 2 left parenthesis straight g right parenthesis close square brackets cubed end fraction
space space space space space equals space fraction numerator left parenthesis 2.0 space cross times space 10 to the power of negative 3 end exponent right parenthesis squared over denominator left parenthesis 1.0 space cross times 10 to the power of negative 3 end exponent right parenthesis thin space left parenthesis 3.0 cross times 10 to the power of negative 3 end exponent right parenthesis cubed end fraction
space space space space space equals fraction numerator 4.0 space cross times space 10 to the power of negative 6 end exponent over denominator 27.0 space cross times space 10 to the power of negative 12 end exponent end fraction space equals 0.149 space cross times space 10 to the power of 6
space space space space space almost equal to space 1.5 space cross times space 10 to the power of 6
Since Qc > Kc, so reaction will go in the left direction and ammonia decomposes into hydrogen and nitrogen.