What will be the initial rate of a reaction if its rate constant is 10^-3 min^-1. and the concentration of
reactant is 0.2 mol dm^-3. How much of reactant will be converted into products in 200 min
a) 2 x 10^-4M min^-1, 18%
b) 10^-4M min^-1, 16%
c) 10^-3M min^-1, 18%
d) 10^-3M min^-1, 17%
please answer only if you know
Answers
EXPLANATION.
Rate Constant = 10⁻³ Min⁻¹.
Concentration of reactant = 0.2 mol dm⁻³.
Time = 200 minutes.
As we know that,
Initial Rate = k[A].
⇒ 10⁻³ [ 0.2].
⇒ 10⁻³ [2/10].
⇒ 2 X 10⁻⁴ M min⁻¹.
From First Order Reaction.
Formula ⇒ k = 2.303/t ㏒(a/a - x).
Put the values in equation, we get.
⇒ 10⁻³ = 2.303/200 ㏒(a/a - x).
⇒ (200)(10⁻³) = 2.303 ㏒(a/a - x).
⇒ 200/(10³)(2.303) = ㏒(a/a - x).
⇒ 200/2303 = ㏒(a/a - x).
⇒ 0.0868 = ㏒(a/a - x).
Antilog,
⇒ 1.22 = ㏒(a/a - x).
⇒ 1.22(a - x) = a.
⇒ 1.22a - 1.22x = a.
⇒ 1.22a - a = 1.22x.
⇒ 0.22a = 1.22x.
⇒ x = 0.22a/1.22.0.
⇒ x = 0.180a.
⇒ 0.180a/a X 100.
⇒ 18%.
Hence Option [A] is correct answer.
MORE INFORMATION.
Zero order reaction.
Integrated law ⇒ A = A₀ - kt.
T(1/2) = A₀/2k.
T(f) = A₀/k.
First order reaction.
Integrated rate law ⇒ t = 2.303/k ㏒(A₀/A).
T(1/2) = 0.69/k.
Hi dear user!!
Rate Constant = 10⁻³ Min⁻¹.
Concentration of reactant = 0.2 mol dm⁻³.
Time = 200 minutes.
As we know that,
Initial Rate = k[A].
⇒ 10⁻³ [ 0.2].
⇒ 10⁻³ [2/10].
⇒ 2 X 10⁻⁴ M min⁻¹.
From First Order Reaction.
Formula ⇒ k = 2.303/t ㏒(a/a - x).
Put the values in equation, we get.
⇒ 10⁻³ = 2.303/200 ㏒(a/a - x).
⇒ (200)(10⁻³) = 2.303 ㏒(a/a - x).
⇒ 200/(10³)(2.303) = ㏒(a/a - x).
⇒ 200/2303 = ㏒(a/a - x).
⇒ 0.0868 = ㏒(a/a - x).
Antilog,
⇒ 1.22 = ㏒(a/a - x).
⇒ 1.22(a - x) = a.
⇒ 1.22a - 1.22x = a.
⇒ 1.22a - a = 1.22x.
⇒ 0.22a = 1.22x.
⇒ x = 0.22a/1.22.0.
⇒ x = 0.180a.
⇒ 0.180a/a X 100.
⇒ 18%.
Hence Option [A] is correct answer.
MORE INFORMATION.
Zero order reaction.
Integrated law ⇒ A = A₀ - kt.
T(1/2) = A₀/2k.
T(f) = A₀/k.
First order reaction.
Integrated rate law ⇒ t = 2.303/k ㏒(A₀/A).
T(1/2) = 0.69/k.