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In the presence of vanadium oxide, SO2 (g) reacts with an excess of oxygen to give SO3 (g): SO2 (g) + ½ O2 (g) → SO3 (g) This reaction is an important step in the manufacture of sulfuric acid. It is observed that tripling the SO2 concentration increases the rate by a factor of 3, but tripling the SO3 concentration decreases the rate by a factor of 1.7 ≈ √3. The rate is insensitive to the O2 concentration as long as an excess of oxygen is present. (a) Write the rate expression for this reaction, and give the units of the rate constant k. (b) If [SO2] is multiplied by 2 and [SO3] by 4 but all other conditions are unchanged, what change in the rate will be observed?

Respuesta :

Explanation:

SO2 (g) + ½ O2 (g) → SO3 (g)

It is observed that tripling the SO2 concentration increases the rate by a factor of 3

This means the reaction is first order with respect to SO2.

The rate is insensitive to the O2 concentration as long as an excess of oxygen is present.

This means that the reaction is Zero order with respect to O2

(a) Write the rate expression for this reaction

Rate = k [SO2]¹[O2]⁰ = k[SO2]

In first order reactions, the reaction rate is directly proportional to the reactant concentration and the units of first order rate constants are 1/sec

Unit for rate constant = 1/sec

(b) If [SO2] is multiplied by 2 and [SO3] by 4 but all other conditions are unchanged, what change in the rate will be observed?

The rate would be unchanged because;

If [SO2] is multiplied by 2; the reaction rate would increase by a factor of 2.

If [SO3] is multiplied by 4; the reaction rate would decrease by a factor of √4 = 2

The increase would cancel the increase.

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