For the reaction X rightarrow products, a plot of [X] vs. time gives a straight

Question

For the reaction X rightarrow products, a plot of [X] vs. time gives a straight line with slope =.0277Ms^-1. After 11.2 s, [X] = 2.23 M. Calculate [X] at t = 0 s. 3.84M 3.29 M 2.77 M 2.54M 2.46 M The second-order reaction A rightarrow products has an initial rate of 0.0335 M s^-1 when [A]_o = 0.440 M. How long will it take for the concentration of A to decrease to 50.0% of its initial value? 1.27 s 4.01 s 5.09 s 13.1 s 29.9 s Which one of the following factors does not affect the solubility of a carbon dioxide gas in a liquid? the temperature of the liquid partial pressure of carbon dioxide above the liquid the identity of the liquid the vapor pressure of the liquid the strength of intermolecular forces between the carbon dioxide and the solvent molecules for the reaction X rightarrow products, a plot of [X] vs. time gives a straight line with slope -0.0277 M s^-1. After 11.2 s. [X] = 2.23 M. Calculate [X] at t = 0 s. -3.84 M^3.29 M 2.77 M 2.54M 2.46 M

Explanation / Answer

18) The plot of [X] vs t gives a straight line with slope = -0.0277 M s-1. The reaction is zero order as can be derived from the integrated rate law below:

-d[X]/dt = k where k = rate constant for the zero order reaction.

==> -d[X] = k.dt

Integrate both sides to get

-{[X]t – [X]0} = k.(t – 0) where [X]0 is the concentration at time t = 0. Therefore,

[X]t – [X]0 = -kt

==> [X]t = [X]0 – kt …..(1)

The reaction is of the form y = mx + c where the slope of the line is m = -k.

Given, slope = -0.0277 M s-1, k = -(slope) = -(-0.0277 M s-1) = 0.0277 M s-1.

Plug in [X]t = 2.23 M and t = 11.2 s and obtain

(2.23 M) = [X]0 – (0.0277 M s-1)*(11.2 s) = [X]0 – 0.31024 M

===> [X]0 = (2.23 + 0.31024) M = 2.54024 M

The answer is (D) 2.54 M (ans).

16) Consider the reaction A -----> Products

is 2nd order in A, i.e., the rate law for the reaction can be written as

-d[A]/dt = k[A]2 where k is the 2nd order rate constant …..(1)

Plug in the value of the rate to determine k.

-(0.0335 M s-1) = k*(0.440 M)2

===> k = -(0.0335 M s-1)/(0.1936 M2) = -0.1730 M-1 s-1.

Re-arrange expression 1 as

-d[A]/[A]2 = k.dt

Integrate and obtain

-(1/[A]t – 1/[A]0) = kt

==> 1/[A]t = 1/[A]0 – kt

Given [A]t = 50%*[A]0 = 0.5*[A]0 = 0.5*(0.440 M) = 0.220 M; plug in values to determine t.

1/(0.220 M) = 1/(0.440 M) – (-0.1731 M-1 s-1)*t

===> (0.1731 M-1 s-1)*t = 1/(0.220 M) – 1/(0.440 M) = (1/0.440) M-1

===> (0.1731 M-1 s-1)*t = 2.273 M-1

===> t = (2.273 M-1)/(0.1731 M-1 s-1) = 13.13 s 13.1 s

Ans: (D) 13.1 s

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