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salantis [7]
3 years ago
15

The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if

the initial concentration of a was 4.00 m, what was the concentration of a (in m) after 180.0 min?
Chemistry
2 answers:
algol133 years ago
8 0
<span>Answer: 0.199M
</span><span />

<span>Explanation:


</span><span>1) An initial clarification: the unit of the concentration is M (molarity) and not m as it was written in the question (m is used for molality, and that is other unit of concentration).
</span>
<span /><span /><span>
2) Second order reaction means that the rate of reaction is given by:
</span><span />

<span>r = - d[A]/dt = [A]²</span>
<span /><span /><span>
3) By integration you get:
</span><span />

<span>1 / [A] - 1[Ao] = kt
</span><span />

<span>=> 1 / [A] = 1 / [Ao] + kt


</span><span>4) Plug in the data; [Ao] = 4.00M; k = 0.0265 (M⁻¹) (min)⁻¹; t = 180 min</span>
<span /><span>
=> 1 / [A] = 1 / 4.00M + 0.0265 (M⁻¹)(min⁻¹) (180min) = 5.02 (M⁻¹)
</span><span />

<span>=> [A] = 1 / (5.02(M⁻¹) = 0.199 M
</span><span>
</span>


katovenus [111]3 years ago
7 0

The integrated rate law for a second-order reaction is given by:

\frac{1}{[A]t} =   \frac{1}{[A]0} + kt

where, [A]t= the concentration of A at time t,

[A]0= the concentration of A at time t=0

<span>k =</span> the rate constant for the reaction


<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min


Hence, \frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)

<span>                                        = 4.858</span>

<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>

<span>
</span>

<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M

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