Question

For the reactions of I, II and III orders, $K_1=K_2=K_3$ when concentrations are expressed in mol $litr{e}^{-1}$ . What will be the relation in $K_1,K_2,K_3$ , if concentrations are expressed in mol/ml ?

• A. $K_1=K_2=K_3$
• B. $K_1=K_2\times {10}^{-3}=K_3\times {10}^{-6}$
• C. $K_1=2K_2=K_3$
• D. $2K_1=3K_2=4K_3$

Answer 1

Answer: (B)

$K_1=K_2\times {10}^{-3}=K_3\times {10}^{-6}$

The rate law expression is $rate=K[Reactant{]}^m.$
Let ${}^{\prime }{a}^{\prime }mollitr{e}^{-1}$ be the concentration of reactant.
The rate law expression for first order reaction $r_1=K_1{\left[a\right]}^1$......(1)
If ${}^{\prime }{a}^{\prime }molm{l}^{-1}$ be the concentration of reactant.
Then concentration in $mollitr{e}^{-1}=a\times {10}^3.$
The rate law expression becomes $r_1=K_1[a\times {10}^3{]}^1$......(2)
Similarly for second order reaction, $r_2=K_2\times {10}^6[a{]}^2$And for third order reaction, $r_3=K_3\times {10}^9[a{]}^3.$
But $K_1=K_2=K_3.$
Hence, $\frac{K_1}{{10}^3}=\frac{K_2}{{10}^6}=\frac{K_3}{{10}^9}$
or $K_1=K_2\times {10}^{-3}=K_3\times {10}^{-6}.$