Question

Review | Constants | Periodic Table The iodate ion, IO3-, is reduced by sulfite, SO32-, according to the following net ionic equation: Part A 103 - (aq) + 3SO32- (aq)-I- (aq) + 3SO42- (aq) The reaction is found to be first order in 103-, first order in SO32-, and first order in H+. If [103] = x, [SO32-] = y, and fH+1 = 2, what is the rate law for the reaction in terms of x, y, and z and the rate constant k? Express the rate in terms of k, x, y, and z (e.g., kxy^3z^2). ► View Available Hint(s) rate = kxyz

Need help on part B and C

Answer 1

Part b

pH = 5.50

Then, [H⁺] = 10^-5.50 = 3.16×10^-6 M

pH = 2.00

Then, [H⁺] [ = 10^-2 M

Rate law

R = K [ IO₃^-][SO₃^-][H⁺]

Now,

When pH = 5.50

Rate₁ = K [IO₃^-][SO₃^-2](3.16×10^-6)

At pH = 2

Rate₂ = K [IO₃^-][SO₃^-2](10^-2)

Then, Rate₂/Rate₁

= $\frac{10^{-2}}{3.16\times 10^{-6}}$

= 0.316 × 10⁴

= 3.16×10³.

= 3.2×10³ ( 2 significant figure).

Part C

H⁺ is involved in the rate law equation , so it is present in the rate-determining state. Catalyst increases the rate of reaction and present in the rate law. Hence , in this reaction H⁺ serves as a catalyst.