Question

In: Chemistry

For each of the following reactions, write down the rate of the reaction in terms of...

For each of the following reactions, write down the rate of the reaction in terms of the appearance of products and disappearance of reactants a) 2 N2O5(g) → 4 NO2(g) + O2(g) b) CH3Cl (g) + 3 Cl2(g) → CCl4 (g) + 3 HCl (g) 2. Given the following balanced equation, determine the rate of reaction with respect to [O2].

2 O3(g) → 3 O2(g)

3. Given the following balanced equation, determine the rate of reaction with respect to [SO3]. If the rate of O2 loss is 3.56 × 10N3 M/s, what is the rate of formation of SO3? 2 SO2(g) + O2(g) → 2 SO3(g)

4. What is the overall order of the following reaction, given the rate law? 2 X + 3 Y → 2 Z Rate = k[X]1[Y]2 What will happen to the Rate of the concentration of Y is doubled?

7. For a particular first order reaction, it takes 120.0 min for the concentration of the reactant to drop to 15% of its initial value. What is the rate constant for this reaction?

8. How many half lives are required for the concentration of reactant to decrease to 25% of its original value?

9. Fluorine 18 undergoes positron emission with a half life of 1.10 × 102 minutes. If a patient is given a 248 mg dose for a PET scan, how long will it take for the amount of fluorineN18 to drop to 83 mg? (Assume that none of the fluorine is excreted from the body.)

10. The second order decomposition of HI has a rate constant of 1.80 × 10N3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?

11. The first order rearrangement of CH3NC is measured to have a rate constant of 3.61 × 10^15 s-1 at 298 K and a rate constant of 8.66 × 10N7 s-1 at 425 K. Determine the activation energy for this reaction.

12. The decomposition of NOCl has k= 9.3 x 10N5 MN1 s N1 at 100. °C and k= 1.0 x 10N3 MN1 s N1 at 130 °C. Calculate the value for the activation energy for the decomposition of NOCl.

13. Consider the reaction: CO(g) + Cl2(g) → COCl2(g) The reaction is first order in CO and 3/2 order in Cl2

a) Write down the rate law for this reaction.

b) What will be the rate of the reaction when [CO] = 0.025 M and [Cl2] = 0.015 M, given that the rate constant is 11 MN3/2sN1

c) Calculate the rate when the concentration of CO is increased to 0.10 M and while the concentration of Cl2 is decreased to 0.0010 M

14. The decomposition of NOBr follows second order kinetics. The rate constant is found to be 0.556 MN1 s N1 . If the initial concentration of NOBr in the container is 0.25 M, how long will it take for the concentration to decrease to 0.025 M?

What is the halfNlife of the decomposition when the initial reactant concentration is 0.25 M?

NOTE: There are some N in questions which are typo

Solutions

Expert Solution

1) a) 2 N2O5(g) → 4 NO2(g) + O2(g)

Rate of appearance and disappearance is =

b) CH3Cl (g) + 3 Cl2(g) → CCl4 (g) + 3 HCl (g)

2) 2 O3(g) → 3 O2(g)

Rate of reaction with respect to O2 =

3) 2 SO2(g) + O2(g) → 2 SO3(g)

Given, rate with respect to O2 =

Rate with respect to SO3 =

4)

2 X + 3 Y → 2 Z

Rate = k[X]1[Y]2

Here 1 and 2 indicates order of the reaction with respect to X and Y, respectively.

Overall order of the reaction = sum of the individual orders = 1 + 2 = 3

Overall order of the reaction = 3

Given condition: If concentration of Y is doubled

In a reaction K wont change with concentration, so k is constant

But Rate changes

Rate1 = k[X]1[Y]2 before Y is doubled

Rate2 = k[X]1[2Y]2 after Y is doubled

divide rate 2 with rate 1

So if concentration of Y is doubled rate changes 4 times the initial rate

7)

Given, time = 120 min

Let initial concentration = Ao = 100%

final concentration = At = 15%

Formula

Rate constant, k =

Rate constant, k= 0.0158 min-1

---------------------------------------

8)

100% -------- 50% -----------25%

in first half life, 100% sample is converted to 50%

In second half life 50 % sample is converted to 25%

So two half lifes are required

9) given, half-life = 1.10 x 102 min

Rate constant = k =

K = 0.0063 min-1

We are asked to find time, t required to get 83 mg from 248 mg dose

formula,

= 1.74 x 102 min

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