Question

hypophosphate ion(H2PO2-) decomposes in alkaline solution as a result of the folowing reaction.

H2PO2-+OH-----HP032-+H2

the kinetic of this reaction were bstudied at 100 deg c.the following table lists the initial concentations of the reactants(from three seperate experiment)and the concentrationof one of the product(HPO32- after the reaction has run for 30 seconds

	[H2PO2]I	[OH-]I	[HPO32-]30S	RATE
EXP1	0.10M	1.0M	1.6*10-5M
EXP2	0.30M	1.0M	4.8*10-5M
EXP3	0.10M	2.0M	6.4*10-5M

a)complete the chart by calculating the initial rate for each expriment (don not forget to include units).hints what is initial concentration of the product? what is the delta t?how is the rate related to product change in concentration?

b) determine the rate law for this reaction.

c)determine the value and units of the rate constant.

d)calculate the rate of reaction if [H2PO2-]I=0.20M and[OH-]I=0.30M

Answer 1

(a) Rate of reaction = change in concentration of reactant or product / Time interval

Rate = [HPO32-] /t

t = 30 - 0 = 30sec

For exp 1, Rate = 1.6 x 10^-5 - 0/ 30 = 0.053 x 10^-5

Exp 2, Rate = 4.8 x 10^-5 - 0/ 30 = 0.16 x 10^-5

Exp 3, Rate = 6.4 x 10^-5 - 0/ 30 = 0.213 x 10^-5

(b) Rate law

Let rate law: R = k[H2PO2-]^x [OH-]^y

We have to calculate value x and y

Consider exp 1 and 2

0.053 x 10^-5 = k[H2PO2-]^x [OH-]^y = k[0.1]^x [1.0]^y   (For exp 1)

0.16 x 10^-5 = k[H2PO2-]^x [OH-]^y = k[0.30]^x [1.0]^y   (For exp2)

Divide these two equation, we get

(3.01)¹ = (3)^x

x = 1

Now consider exp1 and 3

0.053 x 10^-5 = k[H2PO2-]^x [OH-]^y = k[0.1]^x [1.0]^y   (For exp 1)

0.213 x 10^-5 = k[H2PO2-]^x [OH-]^y = k[0.1]^x [2.0]^y (For exp 3)

Dividing these two equation

4.01 = 2^y

(2)² = 2^y

Hence y = 2

Rate law R = k[H2PO2-]¹ [OH-]²

(c) k = R / [H2PO2-]¹ [OH-]²

R = moles L-1 s-1

K = moles L^-1 s^-1 / (moles L^-1)³

K = L² mol⁻² s⁻¹.

(d) First calculate rate constant k, for the given reaction

From exp 1, 0.053 x 10^-5 = k[H2PO2-]¹ [OH-]² = k[0.1]¹ [1.0]²

k = 0.053 x 10^-5 / [0.1]¹ [1.0]² = 0.53 x 10^-5L² mol⁻² s⁻¹.

R = k[H2PO2-]¹ [OH-]²

R = k(0.20) (0.30)²

   = 0.53 x 10^-5 (0.018) = 0.00954 x 10^-5 mol L-1 s-1