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Question

Mixing an acidic solution containing iodate, IO3 - (aq), ions with another solution containing I- (aq) ions begins a reaction that proceeds, in several steps, to finally produce molecular iodine as one of the products as shown in the following reaction. IO3 - (aq) + 5 I- (aq) + H+ (aq) Æ 3 I2(aq) + 3H2O(l) The data in Table 3 are obtained for rate of production of iodine (I2)

Table 1 Concentrations of Reactants and Rate of Product Production

IO3−(aq)   + 5 I− (aq) + 6 H+ (aq)   → 3 I2 (aq) + 3 H2O(l)



Trial	Initial [IO3−] mol/L	Initial [I−] mol/L	Initial [H+] mol/L	Rate of production of iodine (mol/(L●s))
1	0.10	0.10	0.10	r1 = 5.0 X 10 −4
2	0.20	0.10	0.10	r2 = 1.0 X 10 −3
3	0.10	0.30	0.10	r3 = 1.5 X 10 −3
4	0.10	0.30	0.20	r4 = 6.0 X 10 −

1. What is the rate equation for this reaction?
2. What will the rate of reaction be when [IO3− ] =0.20 mol/L [I−] = 0.40 mol/L, and [H+] = 0.10 mol/L?

Answer 1

Comparing trials 1 and 2, doubling the IO3^- concentration doubles the reaction rate, suggesting the reaction is first order in IO3^-.

Comparing trials 1 and 3, tripling the I^- concentration triples the reaction rate, again suggesting first order in I^-.

Comparing trials 3 and 4, doubling the H⁺ concentration results in a 4 fold increase in reaction rate, suggesting second order in H⁺.

The rate equation is Rate = k [IO3^-] [I^-] [H⁺]²

To find k, substitute the concentrations for one of the trials and solve for k.

Using trial 4, k = 6.0 x 10^-3 / 0.10 x 0.30 x 0.20² = 5

The rate for the given concentrations is R = 5 x 0.20 x0..40 x 0.10² = 0.004 mol/(L●s