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Sati [7]
2 years ago
11

4. a solution that is 0.025M in HCOOH and 0.025M in NaCOOH

Chemistry
1 answer:
saveliy_v [14]2 years ago
4 0

KOH added : 7.5 ml

<h3>Further explanation</h3>

Buffer solution of weak acid HCOOH and strong base KOH

Reaction

initial = 100 ml 0.1 M HCOOH = 10 ml mol HCOOH, and x mlmol of KOH

KOH + HCOOH ⇒ COOHK + H₂O

  x           10

  x            x                 x              x

  -            10-x             x              x

[HCOO - ] = 3[HCOOH]

\tt \dfrac{x}{x+100~ml}=3\dfrac{10-x}{x+100}\\\\x=3(10-x)\\\\x=30-3x\\\\4x=30\rightarrow x=7.5~ml

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3 years ago
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For the reaction 2N2O5(g) &lt;---&gt; 4NO2(g) + O2(g), the following data were colected:
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Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

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20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

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At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

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At t = 30 minutes, C = 0.0050 M

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At t = 40 minutes

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At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

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