For a reaction consider the plot of ln k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10-5 s-1, then the rate constant at 500 K is

Question

For a reaction consider the plot of ln k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10-5 s-1, then the rate constant at 500 K is (A) 2 × 10-4 s-1 (B) 10-4s-1 (C) 10-6s-1 (D) 4 × 10-4s-2

Tardigrade Admin · Accepted Answer

We know, ln ( K 2/ K 1)=( E a / R )[(1/ T 1)-(1/ T 2)] Given that, K 1=10-5 ; T 1=400 K ; T 2=500 K ; -( E a / R )=-4606 K Now, 2.303 log ( K 2/10-5)=4606[(1/400)-(1/500)] ⇒ K 2=10-4 s -1

Q. For a reaction consider the plot of $l n k$ versus $1/ T$ given in the figure. If the rate constant of this reaction at $400 K$ is $1 0^{- 5} s^{- 1}$ , then the rate constant at $500 K$ is

$2 \times 1 0^{- 4} s^{- 1}$

$1 0^{- 4} s^{- 1}$

$1 0^{- 6} s^{- 1}$

$4 \times 1 0^{- 4} s^{- 2}$

Q. For a reaction consider the plot of lnk versus 1/T given in the figure. If the rate constant of this reaction at 400K is 10−5s−1, then the rate constant at 500K is

2×10−4s−1

10−4s−1

10−6s−1

4×10−4s−2

We know, <br/>lnK1​K2​​=REa​​[T1​1​−T2​1​]<br/> Given that, <br/>K1​=10−5;T1​=400K;T2​=500K;−REa​​=−4606K<br/> Now, <br/>2.303log10−5K2​​=4606[4001​−5001​]<br/> <br/>⇒K2​=10−4s−1<br/>

Q. For a reaction consider the plot of $l n k$ versus $1/ T$ given in the figure. If the rate constant of this reaction at $400 K$ is $1 0^{- 5} s^{- 1}$ , then the rate constant at $500 K$ is

$2 \times 1 0^{- 4} s^{- 1}$

$1 0^{- 4} s^{- 1}$

$1 0^{- 6} s^{- 1}$

$4 \times 1 0^{- 4} s^{- 2}$