The concentration of a reactant X decreases from 0.1 M to 0.005 M in 40 minutes. If the reaction follows 1 order kinetics, the rate of the reaction when the concentration of X is 0.01 M will be

Question

The concentration of a reactant X decreases from 0.1 M to 0.005 M in 40 minutes. If the reaction follows 1 order kinetics, the rate of the reaction when the concentration of X is 0.01 M will be (A) 1.73 × 10-4M min-1 (B) 3.47 × 10-4M min-1 (C) 3.47 × 10-5M min-1 (D) 7.5 × 10-4 M min -1

Tardigrade Admin · Accepted Answer

k=(2.303/t) log (A0/A)=(2.303/40) log (0.1/0.005) =(2.303/40) log 20=0.075 Rate of reaction when concentration of X is 0.01 M will be 0.075 × 0.01=7.5 × 10-4 M min -1

Q. The concentration of a reactant $X$ decreases from $0.1 M$ to $0.005 M$ in $40$ minutes. If the reaction follows $1$ order kinetics, the rate of the reaction when the concentration of $X$ is $0.01 M$ will be

$1.73 \times 1 0^{- 4} M mi n^{- 1}$

$3.47 \times 1 0^{- 4} M mi n^{- 1}$

$3.47 \times 1 0^{- 5} M mi n^{- 1}$

$7.5 \times 1 0^{- 4} M mi n^{- 1}$

$k = \frac{2.303}{t} lo g \frac{A _{0}}{A} = \frac{2.303}{40} lo g \frac{0.1}{0.005}$

$= \frac{2.303}{40} lo g 20 = 0.075$

Rate of reaction when concentration of $X$ is $0.01 M$ will be

$0.075 \times 0.01 = 7.5 \times 1 0^{- 4} M min^{- 1}$

Q. The concentration of a reactant X decreases from 0.1M to 0.005M in 40 minutes. If the reaction follows 1 order kinetics, the rate of the reaction when the concentration of X is 0.01M will be

1.73×10−4Mmin−1

3.47×10−4Mmin−1

3.47×10−5Mmin−1

7.5×10−4Mmin−1