Question

Question 11 5 pts

Nitrosyl chloride, NOCl, decomposes to NO and Cl2.

2NOCl (g) ----> 2NO (g) + Cl2 (g)

Determine the overall order and rate constant "k" for this reaction from the following data:

[NOCl] (M)

0.10

0.20

0.30

Rate (M/h)

8.0 x 10–10

3.2 x 10–9

7.2 x 10–9

Answer 1

answer
rate = k[NOC]²;

k = 8.0 × 10⁻⁸ L/mol/s;

second order

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Where M

and

rate1 are the concentration and rate of one experiment

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and M2 and rate2     are the concentration and rate of another.

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ratio of the respective rates for the NOCl concentrations.

3.2*10^−9 / 8*10^−10==4  :

this shows that the rate is being quadrupled

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From above

NOCl concentration is doubled, the rate is quadrupled.

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have to determine what it takes in order for the doubling of the concentration to quadruple the rate:

2^x==4

solving for x, we find that x=2,

meaning the NOCl is second order.

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Let us assume Experiment 1 uses .3 M and

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Experiment 2 uses .2 M:

7.2*10^−9 /3.2*10^−9=2.25

= 0.3 /0.2 =1.5

We find that the rate order of NOCl remains two:

1.5^2=2.25

Therefore the rate equation is rate= k [NOCl]^2

s there NOCl is the only reactant, the sum of the exponents of the reactants is two and, therefore, the order for this equation is a second order.

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sing the rate law found above, we can use the rate at a given concentration of NOCl to find the rate constant, k.

8.0⋅10^−10(Mh)=k [0.1M]^2

8.0⋅10^−10(Mh)=k [0.1M]^2

solving for k,

k= 8.0 × 10^⁻⁸ /M⋅hour

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In order to determine the units for k consider the units for the rate and the concentration:

M/h=M^2⋅x

and thus:

x=M/M^2⋅h=1/M⋅h

Molarity is squared due to the rate order of NOCl being two.

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