You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Factors Affecting Rate of Reaction

For the react...

Question

For the reaction
$A + B \to product$
rate law is
rate $= k [A]^{2} [B]$
where $k = 5 \times 10^{- 5} m o l^{- 2} L^{2} m i n^{- 1}$
Determine the time (in minutes) in which concentration of 'A' becomes half of its initial concentration. If initial concentration of A and B are $0.2 M$ and $2 \times 10^{3} M$ respectively.

Open in App

Solution

Since B is in excess, the rate law will become
Rate = $k^{'} [A]^{2}$ where $k^{'} = k [B]$
$\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}} = k^{'} t \frac{1}{[A]_{2} k^{'}} = t_{\frac{1}{2}} t_{\frac{1}{2}} = \frac{1}{0.2 \times 5 \times 10^{- 5} \times 2 \times 10^{3}} = 50 m i n u t e s$

Suggest Corrections

Similar questions

Q. Reaction;

A + B ⟶ C + D

, follows rate law;

r a t e = k {[A]}^{2} {[B]}^{2}

. Starting with initial concentration of

1 M

A

and

B

each, what is the time taken for concentration of

A

to become

0.25 M

?
Given:

k = 2.303 \times 10^{- 2} {s e c}^{- 1}

Q. Reaction;

A + B \to C + D

, follows rate law;

r a t e = k [A]^{\frac{1}{2}} [B]^{\frac{1}{2}}

. Starting with initial conc, of

1 M

A

and

B

, what is the time taken for concentration of

A

of becomes

0.25 M

?
Given:

k = 2.303 \times 10^{- 3} {sec}^{- 1}

Q. For the reaction

A \to B

, rate constant

K = 2 \times 10^{- 4} m o l L^{- 1} S^{- 1}

. If initial concentration of

A

1 m o l L^{- 1}

concentration of

A

after

20

minute is:

Q. Reaction

A + B \to C + D

follows following rate law: rate

= k [A]^{+ \frac{1}{2}} [B]^{\frac{1}{2}} .

Starting with initial conc. of

1 M

A

and

B

each, what is the time taken for concentration of

A

to become 0.25 M?

(Given :

K = 2.303 \times 10^{- 3} s e c^{- 1}

)

Q. For the zero order reaction

A \to B + C

: initial concentration of

A

0.1 M

. If

A = 0.08 M

after

10 minutes

, then its half-life and completion time are respectively :

Join BYJU'S Learning Program

For the reaction A+B→product rate law is rate =k[A]2[B] where k=5×10−5 mol−2 L2 min−1 Determine the time (in minutes) in which concentration of 'A' becomes half of its initial concentration. If initial concentration of A and B are 0.2 M and 2×103 M respectively.

Since B is in excess, the rate law will become Rate = k′[A]2 where k′=k[B] 1[A]t−1[A]0=k′t1[A]2k′=t12t12=10.2×5×10−5×2×103=50 minutes

Since B is in excess, the rate law will become
Rate = $k^{'} [A]^{2}$ where $k^{'} = k [B]$
$\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}} = k^{'} t \frac{1}{[A]_{2} k^{'}} = t_{\frac{1}{2}} t_{\frac{1}{2}} = \frac{1}{0.2 \times 5 \times 10^{- 5} \times 2 \times 10^{3}} = 50 m i n u t e s$