Consider the reaction shown below.
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?
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.
To find the rates of production of P4 and H2, we first need to determine the stoichiometry of the reaction.
From the balanced equation, we can see that for every 4 moles of PH3 consumed, 1 mole of P4 and 6 moles of H2 are produced.
Given that 0.0061 mol of PH3 is consumed each second, we can calculate the rates of production as follows:
Rate of production of P4 = (0.0061 mol PH3/s) x (1 mol P4/4 mol PH3)
Rate of production of P4 = 0.0061/4 mol/s
Rate of production of H2 = (0.0061 mol PH3/s) x (6 mol H2/4 mol PH3)
Rate of production of H2 = 0.0061 x 6/4 mol/s
Simplifying the calculations, we get:
Rate of production of P4 = 0.001525 mol/s
Rate of production of H2 = 0.00915 mol/s
Therefore, the rate of production of P4 in this experiment is 0.001525 mol/s, and the rate of production of H2 is 0.00915 mol/s.
To determine the rates of production of P4 and H2 in the given reaction, we need to use the stoichiometric coefficients of the balanced equation.
According to the balanced equation:
4 PH3(g) → P4(g) + 6 H2(g)
We can see that for every 4 moles of PH3 consumed, 1 mole of P4 is produced, and for every 4 moles of PH3 consumed, 6 moles of H2 are produced. This implies the following molar ratios:
1 mole P4 : 4 moles PH3
6 moles H2 : 4 moles PH3
Based on the given information that 0.0061 mol PH3 is consumed each second, we can calculate the rates of production for P4 and H2 as follows:
Rate of production of P4:
1 mole P4 : 4 moles PH3
Rate of production of P4 : Rate of consumption of PH3
Rate of production of P4 = (1 mole P4 / 4 moles PH3) * (0.0061 mol PH3 / 1 second)
Rate of production of P4 = 0.001525 mol P4/second
Similarly, for H2:
Rate of production of H2:
6 moles H2 : 4 moles PH3
Rate of production of H2 : Rate of consumption of PH3
Rate of production of H2 = (6 moles H2 / 4 moles PH3) * (0.0061 mol PH3 / 1 second)
Rate of production of H2 = 0.00915 mol H2/second
Therefore, the rates of production of P4 and H2 in this experiment are 0.001525 mol P4/second and 0.00915 mol H2/second, respectively.