Question

Part 1.)

A flask is charged with 0.110 mol of A and allowed to react to form B according to the following hypothetical gas-phase reaction.

A(g) B(g)

The following data are collected.

times (s)	0	40	80	120	160
moles of A	0.110	0.070	0.043	0.032	0.024

(a) Calculate the number of moles of B at each time in the table.

0 s
__ mol

40 s
__ mol

80 s
__ mol

120 s
__ mol

160 s
__ mol



(b) Calculate the average rate of disappearance of A for each 40 s interval, in units of mol/s.

0 - 40 s
__ mol/s

40 - 80 s
__ mol/s

80 - 120 s
__ mol/s

120 - 160 mol/s
__ mol/s

Part 2.)

The reaction S2O82-(aq) + 3 I-(aq) 2 SO42-(aq) + I3-(aq) was studied at a certain temperature with the following results:

Experiment	[S2O82-(aq)] (M)	[I-(aq)] (M)	Rate (M/s)
1	0.0309	0.0309	5.99e-06
2	0.0309	0.0618	1.20e-05
3	0.0618	0.0309	1.20e-05
4	0.0618	0.0618	2.39e-05

(a) What is the rate law for this reaction?

Rate = k [S₂O₈^2-(aq)] [I^-(aq)], Rate = k [S₂O₈^2-(aq)]² [I^-(aq)], Rate = k [S₂O₈^2-(aq)] [I^-(aq)]², Rate = k [S₂O₈^2-(aq)]² [I^-(aq)]², Rate = k [S₂O₈^2-(aq)] [I^-(aq)]³, Rate = k [S₂O₈^2-(aq)]⁴ [I^-(aq)]

(b) What is the value of the rate constant?

(c) What is the reaction rate when the concentration of S₂O₈^2-(aq) is 0.0602 M and that of I^-(aq) is 0.0655 M if the temperature is the same as that used to obtain the data shown above? _ M/s

Answer 1

The moles of B are calculated at each instant of time:

0 s: 0 mol

40 s: (0.11 - 0.07) = 0.04 mol

80 s: (0.11 - 0.043) = 0.067 mol

120 s: (0.11 - 0.032) = 0.078 mol

160 s: (0.11 - 0.024) = 0.086 mol

The disappearance rate of A in each interval is calculated:

0 - 40: (0.11 - 0.07) / (40 - 0) = 0.001

40 - 80: (0.07 - 0.043) / 40 = 6.75x10 ^ -4

80 - 120: (0.043 - 0.032) / 40 = 2.75x10 ^ -4

120 - 160: (0.032 - 0.024) / 40 = 2x10 ^ -4

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