In the presence of vanadium oxide, SO2(g) reacts with an excess of oxygen to giv

Question

In the presence of vanadium oxide, SO2(g) reacts with an excess of oxygen to give SO3(g):

SO2(g) + 1/2 O2(g) --> SO3(g) (V2O5 on top of arrow)

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 = sqrt root (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?

For a. the rate expression I got is r = k[SO2][SO3]^-0.5. But are the units of k?

Explanation / Answer

a)

Reaction rates depend on reactant concentrations, temperature, and the presence of catalysts.

As the rate increases three times when [SO2] increased three times, it indicates that it follows first order. Thus,

Rate law = k[SO2]

Units of k = s^-1 as it is a first order reaction.

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b)

If SO2 is increased by two times, the rate also increases by two times. The rate law is independent of concentration of products.

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