For the reversible reaction equilibrium is `N_(2)(g)+O_(2)(g)underset(k_(2))overset(k_(1))(hArr)2NO(g)` `C_(0)=Ce^(-2.1xx10^(3))`for the forward and `C_(0)^(')=C^(')e^(-4.2xx10^(-4t))`for the backward reaction, hence, `K_(c)` for the above equilibrium is (both forward and backward reactions are first order reactions )

Options

A. 5
B. 2
C. 0.5
D. 2

Answer

A

Explanation

For a first oder reaction,
`k=(1)/(t) log_(e) .(C_(0))/(C )`
`:. k_(1)=(1)/(t)log_(e).e^(-2.1xx10^(-3)t)=(-2.1xx10^(-3)t)/(t)`
`=-2.1xx10^(-3)`
`k_(2)=(1)/(t)log_(e).e^(-4.2xx10^(-4)t)`
`=-4.2xx10^(-2)`
`K=(k_(1))/(k_(2))=(-2.1xx10^(-3))/(-4.2xx10^(-4))=(210)/(42)=5`