Question

Rate constant for a particular reaction is $3\times {10}^{-6}$ mole${}^{-2}$ l${}^2$ s${}^{-1}.$ The rate of reaction for only single reactant is found to be $2.4\times {10}^2$ mole l${}^{-1}$ s${}^{-1}$. The equilibrium concentration of the reactant could be:

• A. $0.8\times {10}^4$
• B. $2\times {10}^2$
• C. $0.8\times {10}^2$
• D. $2\times {10}^1$

Answer 1

The correct option is A $0.8\times {10}^4$

According to rate law:

$Rate=k[A{]}^n$

from the units of $k=3\times {10}^{-6}$ mol${}^{-2}$ L${}^2$ s${}^{-1}.$

and $R=2.4\times {10}^{2}mol{L}^{-1}{s}^{-1}$

we get unit of $[A{]}^n=mo{l}^2{L}^{-2}$

since concentration unit is $mol{L}^{-1}$

therefore $n=2$

substitute the values in rate law:

$2.4\times {10}^2=3\times {10}^{-6}\times [A{]}^2$

$[A{]}^2=0.8\times {10}^8$

$\left[A\right]=0.89\times {10}^4$

A is correct.