For the equilibrium the value of the constant, K c is Calculate for the reaction at this temperature. - Wired Faculty

Equilibrium

Question

For the equilibrium
$2 NOC l left parenthesis straight g right parenthesis space space space space rightwards harpoon over leftwards harpoon space space space space 2 NO left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis$
the value of the constant, K_cis $3.75 space cross times space 10 to the power of negative 6 end exponent space at space 1069 space straight K.$ Calculate $straight K subscript straight p$ for the reaction at this temperature.

Solution

We know
$straight K subscript straight p space equals space straight K subscript straight c left parenthesis RT right parenthesis to the power of increment straight n end exponent$ ...(1)

$Here space increment straight n space equals space straight n subscript straight p space minus space straight n subscript straight r space equals space left parenthesis 2 plus 1 right parenthesis minus 2 space equals space 1 space space space space space space space space straight K subscript straight c space equals space 3.75 space cross times space 10 to the power of negative 6 end exponent space space space space space space space space space straight T space equals space 1069$
$straight R space equals 0.0831 space bar space straight L space mol to the power of negative 1 end exponent straight K to the power of negative 1 end exponent$
Putting the values in expression (1), we have

$straight K subscript straight p space equals space left parenthesis 3.75 space cross times space 10 to the power of negative 6 end exponent right parenthesis space left parenthesis 0.0831 right parenthesis cross times left parenthesis 1069 right parenthesis space space space space equals space 0.033$