Mr Cole Chemistry

Introduction to Orders Questions

Introduction to Orders

Rate Equations Worksheet

1. Orders and Equations

1. For the reaction A + 2B → C, the reaction is first order with respect to both A and B. The initial rate of reaction is 4 mol dm-3 s-1.
a) State the overall order of reaction.

Second Order (1 + 1 = 2).

b) Write the rate equation for the reaction.

Rate = k[A][B]

c) State the unit for k.

Units = Rate / ([A][B])
= (mol dm-3 s-1) / (mol dm-3 × mol dm-3)
= mol-1 dm3 s-1

d) State the change in rate if you double the concentration of both A and B.

Rate ∝ [A][B]
Change = 2 × 2 = 4.
The rate will quadruple.

2. For the reaction 2A + 2B → C, the reaction is second order with respect to A, zero order with respect to B and first order with respect to the catalyst D. The initial rate of reaction is 40 mol dm-3 s-1.
a) Write the rate equation for the reaction.

Rate = k[A]2[D]
(Note: B is zero order so it does not appear in the rate equation).

b) State the unit for k.

k = Rate / ([A]2[D])
= (mol dm-3 s-1) / ((mol dm-3)2 × mol dm-3)
= (mol dm-3 s-1) / (mol3 dm-9)
= mol-2 dm6 s-1

c) State the change in rate is you multiply the concentration of A by 10 and divide the concentration of B and D by 6.

Rate ∝ [A]2[D]
Effect of A (×10): Rate increases by 102 = 100.
Effect of B (÷6): Zero order, no effect.
Effect of D (÷6): First order, rate reduces by 6.
Total Change: 100 / 6 = 16.67 times faster.

3. For the reaction 2A + 2B → C, the reaction is first order with respect to A and second order with respect to B. The initial rate of reaction is 2.4 mol dm-3 s-1.
a) State the overall order of reaction.

Order = 1 (A) + 2 (B) = Third Order.

b) Write the rate equation for the reaction.

Rate = k[A][B]2

c) State the unit for k.

k = Rate / ([A][B]2)
= (mol dm-3 s-1) / (mol dm-3 × (mol dm-3)2)
= mol-2 dm6 s-1

d) State the change in rate if you double the concentration of B and half the concentration of A.

Rate ∝ [A][B]2
Change = 0.5 (for A) × 22 (for B)
Change = 0.5 × 4 = 2.
The rate doubles.

4. For the reaction 2A + 2B → C, the reaction is second order with respect to A, zero order with respect to B and first order with respect to the catalyst D. The initial rate of reaction is 40 mol dm-3 s-1.
a) Write the rate equation for the reaction.

Rate = k[A]2[D]

b) State the unit for k.

mol-2 dm6 s-1

c) State the change in rate if you half the concentration of Z and multiply the concentration of A by 4.

(Assuming Z refers to the catalyst D based on context):
Rate ∝ [A]2[D]
Change = 42 (for A) × 0.5 (for D)
Change = 16 × 0.5 = 8.
The rate increases by a factor of 8.

2. Calculating Rate

Calculate the rate of reaction for the following:
a) For the reaction C + D → E, where the value of k is 0.15 mol-1 dm3 s-1, and the rate equation is rate = k[C][D], calculate the rate of reaction at [C] = 1.5 moldm-3 and [D] = 2.0 mol dm-3.

Rate = 0.15 × 1.5 × 2.0 = 0.45 mol dm-3 s-1

b) For the reaction 2C + D → E, where the value of k is 0.25 mol-1 dm3 s-1, and the rate equation is rate = k[C]2[D], calculate the rate of reaction at [C] = 1.2 moldm-3 and [D] = 2.0 mol dm-3.

Rate = 0.25 × (1.2)2 × 2.0
Rate = 0.25 × 1.44 × 2.0 = 0.72 mol dm-3 s-1

c) For the reaction 2C + D → E, where the value of k is 1.75 mol-1 dm3 s-1, and the rate equation is rate = k[C]2[D][Z], calculate the rate of reaction at [C] = 0.4 moldm-3, [D] = 0.5 mol dm-3 and [Z] = 0.5mol dm-3.

Rate = 1.75 × (0.4)2 × 0.5 × 0.5
Rate = 1.75 × 0.16 × 0.25 = 0.07 mol dm-3 s-1

d) For the reaction C + 3D → E, where the value of k is 0.005 mol-1 dm3 s-1, and the rate equation is rate = k[C][D]2[Z]2, calculate the rate of reaction at [C] = 2.5 moldm-3, [D] = 3.0 mol dm-3 and [Z] = 3.2 mol dm-3.

Rate = 0.005 × 2.5 × (3.0)2 × (3.2)2
Rate = 0.005 × 2.5 × 9 × 10.24 = 1.152 mol dm-3 s-1

3. Calculating k

Calculate the following values for k, given the data.
a) For the reaction X + 2Y → Z, where [X] = 1.5 mol dm-3 and [Y] = 1.25 mol dm-3, and the rate of reaction is 5.5 mol-1 dm3 s-1. The rate equation is Rate = k [X][Y]2.

k = Rate / ([X][Y]2)
k = 5.5 / (1.5 × 1.252)
k = 5.5 / (1.5 × 1.5625) = 2.35 mol-2 dm6 s-1

b) For the reaction X + 2Y → Z, where [X] = 0.5 mol dm-3; [Y] = 0.6 mol dm-3; and [A] = 1.2 mol dm-3. The rate of reaction is 4.2 mol-1 dm3 s-1. The reaction was first order with respect to Z and Y, and second order with respect to the catalyst, A.

(Note: Question assumes “1st order w.r.t Z” refers to reactant X, as [X] is provided).

Rate = k[X][Y][A]2
k = Rate / ([X][Y][A]2)
k = 4.2 / (0.5 × 0.6 × 1.22)
k = 4.2 / (0.3 × 1.44)
k = 4.2 / 0.432 = 9.72 mol-3 dm9 s-1

c) For the reaction X + 2Y → Z, where [X] = 1.5 mol dm-3 and [Y] = 1.25 mol dm-3, and the rate of reaction is 1.6 mol-1 dm3 s-1. When the concentration of X was increased to 3.0 mol dm-3 (but [Y] was kept the same), the rate was 3.20 mol-1 dm3 s-1. When the concentration of Y was increased to 2.50 (but [X] remained at 1.5 mol dm-3) the rate increased to 3.2 mol-1 dm3 s-1.

Step 1: Determine Orders
Compare Exp 1 ([X]=1.5, Rate=1.6) and Exp 2 ([X]=3.0, Rate=3.2).
[X] doubles, Rate doubles → 1st Order w.r.t X.

Compare Exp 1 ([Y]=1.25, Rate=1.6) and Exp 3 ([Y]=2.50, Rate=3.2).
[Y] doubles, Rate doubles → 1st Order w.r.t Y.

Step 2: Calculate k
Rate = k[X][Y] → k = Rate / ([X][Y])
k = 1.6 / (1.5 × 1.25)
k = 1.6 / 1.875 = 0.853 mol-1 dm3 s-1

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