# Rates Solution Page 4

Remedial Problem Set Part B1 Solution

1. (a) The law of mass action states The rate of reaction is directly proportional to the product of the concentrations of the reactants raised to their coefficients

aA + bB ® cC + dD Rate = k[A]a [B]b

(b) Does the law apply to reactions that involve more than one step? Explain.

Rate law is determined experimentally. If the order from the rate law is different from the order of the reaction from the law of mass action, we can presume that the reaction is not a one-step reaction. When it is not a one-step reaction, it suggests that the reaction takes place through more than one step, and the slowest step in the elementary step suggested is the rate determining step.

In simple terms if the rate law order is different from the molecularity of the reaction the reaction takes place through more than one step.

1. The following are one-step reactions. For one step reactions Rate law is equal to Law of mass action

(1) N2O5 ® 2NO2 + ½ O2 (2)H2 + I2 ® 2HI (3) 2A + B ® C + D (all are gases)

Complete the following table

 Reaction Rate law Overall order Units of k 1 Rate = k [N2O5] 1 Time-1 2 Rate = k[H2] [I2] 1+1 = 2 L mol-1 time -1 3 Rate = k[A]2 [B] 2+1 = 3 L2 mol-2 time-2
1. The reaction HI + CH3I ® CH4 + I2 (all are gases) Is known to be first order with respect both reactants. The rate constant at 50oC is 2.3 x 10-3 litre/mol-min. Starting with equal concentrations of HI and CH3I, each being 0.200 mol/litre, calculate

(a) the concentration of HI remaining after 100 min. assuming that the reaction rate is constant over that period of time (Ans: 0.1908 mol/L)

Rate = k[HI] [CH3I] Substituting values we get Rate = 2.3×10-3 molL-1min-1 [0.2M][0.2M]

Rate = 9.2 x 10-5 molL-1 min-1

Concentration decrease after 100 minutes if found = 100 min x 9.2 x 10-5mol L-1 min-1 = .0092 mol L‑1

After 100 minutes amount of HI remaining will be initial concentration – amount reacted

HI remaining = [0.2 – .0092] = 0.1908 molL-1

(b) the concentration of I2 produced after 200 min (Ans 0.0184 mol/L)

Amount of Iodine or I2 after 200 minutes is = 200 min x 9.2 x 10-5mol L-1 min-1 = 0.0184 mol L‑1

Time times rate gives you the answer.

[I2 ] produced after 200 minutes = 0.1908 molL-1

1. For the following reaction 3A + B ® D + F

(a) What is the rate equation predicted by the Law of Mass Action?

Rate = k[A]3 [B]

(b) What is the overall order? Order n = 3+1 = 4

(c) What is the probability that this reaction has a mechanism? Explain.

For higher molecularity, as in this case which is 4, it is quite unlikely that all the four reactants can come together or collide at the same time and undergo a single step reaction. Collision is unlikely to happen for the 4 particles. Hence it is likely that the experimentally rate law will be different and lower. For that reason we would assume that a mechanism is involved.

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