The rate constant for the reaction, $$2{N_2}{O_5} \to 4N{O_2} + {O_2},\,{\text{is}}\,3.0 \times {10^{ - 5}}{\sec ^{ - 1}}.$$ If the rate is $$2.40 \times {10^{ - 5}}\,{\text{mol}}\,{\text{litr}}{{\text{e}}^{ - 1}}\,{\sec ^{ - 1}},$$ then the concentration of $${N_2}{O_5}$$ ( in $${\text{mol}}\,{\text{litr}}{{\text{e}}^{ - 1}}$$ ) is
A.
1.4
B.
1.2
C.
0.04
D.
0.8
Answer :
0.8
Solution :
TIPS/Formulae :
Find the order of reaction and then use appropriate equation.
As unit of $$k$$ is $${\sec ^{ - 1}}.$$ reaction is of first order,
$$\eqalign{
& r = k\left[ {{N_2}{O_5}} \right]; \cr
& \therefore \,\left[ {{N_2}{O_5}} \right] = \frac{{2.4 \times {{10}^{ - 5}}}}{{3 \times {{10}^{ - 5}}}} \cr
& = 0.8\,mol/L \cr} $$
Releted MCQ Question on Physical Chemistry >> Chemical Kinetics
Releted Question 1
If uranium (mass number 238 and atomic number 92) emits an $$\alpha $$ -particle, the product has mass no. and atomic no.