The rate of the reaction A right arrow B when [A]= 3.5 M is 1.6*10^-2Ms^-1. If t
ID: 950278 • Letter: T
Question
The rate of the reaction A right arrow B when [A]= 3.5 M is 1.6*10^-2Ms^-1. If the reaction is second order in A, the rate constant is - M^-1s^-1. 1.3 2.6 26 0.13 The reaction 4PH^3(g) right arrow P^4(g)+6H^2(g) is a first-order reaction with a half-life of 35s. The rate constant of the reaction is s^-1. 1.12*10^-4 1.98 x 10^-2 1.98 x 10^-3 3.16 x 10^-2 The rate constant of the second-order reaction 2NOB_r(g) right arrow 2NO(g)+Br_2(g) is 0.80 M^1s^-1. Starling with a concentration of NOBr of 0.086 M, the concentration of NOBr alter 22 seconds is 0.022 M 0.076 M 0.062 M 0.034 M. The rate constant of the reaction A right arrow product is 0.54M^-1s^-1. The time taken for the concentration of A to decrease from 0.62 M to 0.28 M is seconds. 6.3 21 3.6 15 The rate constant of a reaction is 4.60*10^-4 s^-1 at 350degreeC. The temperature at which the rate constant will be 8.80* 10^-4 s^-4 is. [The activation energy is 104 kJ mol^-1]. 461degreeC 371degreeC 417degreeC 392degreeC The rate constant for a reaction at 305.0 K is two times the rate constant at 295.0 K. The activation energy of the reaction is kJ mol^-1. 36.2 21.7 72.9 51.8Explanation / Answer
Answer 5:
Given reaction is second order in A,
Hence rate = k[A]2
Given concentration of A = 0.35 M
and rate = 1.6x10-2
putting the values, we get,
k= 1.6x10-2/(0.35)2
k = 0.13
Answer 6:
Given t1/2 = 35 sec
For the reaction to be first order the half life comes out to be,
t1/2 = 0.693/k
or, k = 0.693/t1/2
= 0.693/35
= 1.98x10-2 sec-1
Answer 7:
Given rate constant, k = 0.80 M-1sec-1
and initial concentration [Ao] = 0.086 M
we have for second order reaction,
kt = 1/[A] - 1/[Ao]
after 22 seconds,
0.80x22 = 1/[A] - 1/0.086
1/[A] = 17.6 + 11.62
1/[A] = 29.22
[A] = 0.034 M
Answer 8:
Given rate constant = 0.54 M-1 sec-1
[Ao] = 0.62 M
[A] = 0.28 M
As it is clear from the units of rate constant that the reaction is second order.
Hence as the reation, i.e
kt = 1/[A] -1/[Ao]
0.54x(t) = 1/0.28 - 1/0.62
t = (3.57-1.612)/0.54
t = 1.958/0.54
= 3.6 sec
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