part a a reaction rate can be measured relative to either the rate of reactant l

Question

part a

a reaction rate can be measured relative to either the rate of reactant loss or the rate of product increase. Depending on which species concentration is followed, the possible expressions for the rate of this reaction are: - [02] Rate [H20] Regardless of which species' concentration is followed the following must be true Rate- The rate expressions are always set up to show a positive value relative to the species with the slowest concentration change. Stoichiometric coefficients are used to determine the actual rates for the other species Why does must a rate expression based on reactant concentration change contain a negative sign but a rate expression based on product concentration change is positive? (Hint: what happens to the concentration of each species as the reaction proceeds?) Since reactant concentrations increase, the change in concentration is positive, thus it must be negated to obtain a negative rate None of the other answers are applicable Since reactant concentrations decrease, the change in concentration is positive, thus it must be negated to obtain a negative rate. Since reactant concentrations decrease, the change in concentration is negative, thus it must be negated to obtain a positive rate

Explanation / Answer

Part a.

Last option (Since reactant concentration decreases, the change in concentration is negative, thus it must be negated to obtain a positive rate.) is correct.

Rate = concentration/ time

For a reaction

2H2 + O2 -----> 2H2O

Rate =1/2[H2]/t = [O2]/t = 1/2[H2O]t

The concentration of a reactant H2 and O2 always decreases with time, so [H2] and [O2] are both negative. Since negative rates do not make much sense, to make the rate come out positive, rates expressed in terms of a reactant concentration are always preceded by a minus sign. The concentration of a product H2O always increases with time, so [H2O] is positive.

Part b.

1st option (Hydrogen is used and water is produced at twice the rate of oxygen is used, hence the factor produces the necessary equality in expression.) is correct.

Reaction stoichiometry shows, that 1 mol of hydrogen with 2 moles of oxygen to produce 2 moles of water.

So, it is clear that [H2] decreases twice as rapidly as [O2] and [H2O] is increase twice as rapidly as [O2].So in order to avoid ambiguity in rate expression, it is customary to divide each change in concentration by the appropriate coefficient.

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