Effect of Doubling Reaction Vessel Volume on Reaction Rate

The concentration of reactants in a gas-phase reaction plays a crucial role in determining the reaction rate. This article explores how the doubling of the volume of the reaction vessel affects the rate of the reaction:

2CO O2 → 2CO2

First, it is important to understand that concentration is defined as moles of solute per unit volume. Therefore, if the volume of the reaction vessel is doubled, the concentration of the reactants, CO and O2, will be halved. If the reaction follows a rate law of the form:

Rate k[CO]^a [O2]^b

where a and b are the orders of the reaction with respect to CO and O2 respectively. The new rate after doubling the volume would be:

[CO] → [ CO ] / 2

[O2] → [ O2 ] / 2

Thus, the new rate becomes:

Ratenew k[ CO / 2 ]a[ O2 / 2 ]b

This simplifies to:

Ratenew k[CO]a[O2]b / 2ab

So the reaction rate decreases by a factor of 2ab.

Conclusion

Based on the reaction orders:

If the reaction is first order with respect to both CO and O2, i.e., a 1 and b 1, the rate decreases by a factor of 211 4. If a b 3, the rate would decrease by a factor of 23 8.

Without knowing the specific orders of the reaction, the exact factor by which the rate decreases is dependent on the reaction orders.

For summary conclusions using the given options:

Assuming the simplest case, first order for both reactants, the rate would decrease by a factor of 4, which is not an available option in the provided list. For a third-order reaction, the rate would decrease by a factor of 8, which corresponds to option d (reduce by 8 times).

Detailed Analysis

When the volume of the container doubles, the concentrations are halved, leading to a decrease in the reaction rate. However, the assumption of a one-step process and the rate equation is crucial to determining the exact factor of decrease.

If this reaction is assumed to be a one-step process, the rate equation would be:

Rate k[CO]2[O2]1

Halving the concentrations yields:

Rate2 k[(1/2)CO]2[(1/2)O2]1

Which simplifies to:

Rate2 k(1/8)[CO]2[O2]

This equals:

Rate2 1/8 Rate

Thus, only with this assumption can we conclude that option d (reduce by 8 times) is correct.