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Equilibrium

Question
CBSEENCH11006393

Calculate the value of equilibrium constant for the reaction:
straight N subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space space space rightwards harpoon over leftwards harpoon space space space 2 NO subscript 2 left parenthesis straight g right parenthesis semicolon space space space increment subscript straight r straight H to the power of 0 space equals space minus ve

There is 10.0 mol of N2, 14·0 mol of O2 and 0·2 mol of NO2 present at equilibrium in a 3·0L vessel at 298K.What will be the effect of increased temperature on the equilibrium constant?

Solution
The reversible reaction is
straight N subscript 2 left parenthesis straight g right parenthesis space plus space 2 straight O subscript 2 left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space space 2 NO subscript 2 left parenthesis straight g right parenthesis semicolon space space space increment subscript straight r straight H to the power of 0 space equals space minus ve
The equilibrium constant Kc is given by 
                     straight K subscript straight c space equals space fraction numerator open square brackets NO subscript 2 close square brackets squared over denominator open square brackets straight N subscript 2 close square brackets space open square brackets straight O subscript 2 close square brackets squared end fraction space space space space space space space space space space.... left parenthesis 1 right parenthesis
     open square brackets straight N subscript 2 close square brackets space equals space fraction numerator 10.0 space mol over denominator 3.0 space straight L end fraction space equals space 3.333 space mol space straight L to the power of negative 1 end exponent
open square brackets straight O subscript 2 close square brackets space equals space fraction numerator 14.0 space mol over denominator 300 space straight L end fraction space equals space 4.666 space mol space straight L to the power of negative 1 end exponent
open square brackets NO subscript 2 close square brackets space equals space fraction numerator 0.2 space mol over denominator 3.0 space straight L end fraction space equals space 0.666 space mol space straight L to the power of negative 1 end exponent
Putting the values in expression (1), we have,
     straight K subscript straight c space equals space fraction numerator left parenthesis 0.666 space mol space straight L to the power of negative 1 end exponent right parenthesis squared over denominator left parenthesis 3.333 space mol space straight L to the power of negative 1 end exponent right parenthesis end fraction space left parenthesis 4.666 space mol space straight L to the power of negative 1 end exponent right parenthesis
space space space space equals space fraction numerator left parenthesis 0.666 right parenthesis squared over denominator left parenthesis 3.333 right parenthesis thin space left parenthesis 4.666 right parenthesis end fraction mol to the power of negative 1 end exponent space straight L
space space space space equals space space 6.1 space cross times 10 to the power of negative 5 end exponent space mol to the power of negative 1 end exponent straight L
It is an exothermic reaction. According to Le-Chatterlier's principle, an increase in temperature would shift the equilibrium to left and the value of Kc should decrease.