What is the rate of production of d in mol L-1s-1 if the rate of consumption of a is 0.230 mol L-1s-1 in the reaction 3 A (aq) + 4 B (aq) → C (aq) + 4 D (aq)?

To find the rate of production of D, we first need to understand the stoichiometry of the reaction. The given reaction is:

3 A (aq) + 4 B (aq) → C (aq) + 4 D (aq)

From the balanced equation, we can see that for every 3 moles of A that are consumed, 4 moles of D are produced. This gives us a stoichiometric relationship that we can use to relate the rate of consumption of A to the rate of production of D.

The rate of consumption of A is given as 0.230 mol L-1s-1. We can express this in terms of the rate of production of D using the stoichiometric coefficients. The relationship is as follows:

Rate of production of D = (4/3) * Rate of consumption of A

Now plug in the known rate of consumption of A:

Rate of production of D = (4/3) * 0.230 mol L-1s-1

Rate of production of D = 0.307 mol L-1s-1

Thus, the rate at which D is produced in this reaction is 0.307 mol L-1s-1.

More Related Questions