Consider the reaction shown below.

Question

4 PH3(g) P4(g) + 6 H2(g)
If, in a certain experiment, over a specific time period, 0.0061 mol PH3 is consumed in a 1.9-L container each second of reaction, what are the rates of production of P4 and H2 in this experiment?

Answer 1

We can use the stoichiometry of the balanced chemical equation to determine the rates of production of P4 and H2.

For every 4 moles of PH3 consumed, 1 mole of P4 is produced, and 6 moles of H2 are produced. So, we can set up the following relationships:

Rate of consumption of PH3 : Rate of production of P4 : Rate of production of H2 = 4 : 1 : 6

The rate of consumption of PH3 is 0.0061 mol/s. Thus, we can write the following proportions:

0.0061 mol/s (PH3) / 4 = x mol/s (P4) / 1 = y mol/s (H2) / 6

Solving for x and y:

1. x = 0.0061 mol/s * (1 mol P4 / 4 mol PH3)
x = 0.0061 mol/s * (1/4)
x = 0.001525 mol/s (P4)

2. y = 0.0061 mol/s * (6 mol H2 / 4 mol PH3)
y = 0.0061 mol/s * (3/2)
y = 0.00915 mol/s (H2)

So, the rates of production of P4 and H2 in this experiment are 0.001525 mol/s and 0.00915 mol/s, respectively.