Help with Chem rate of reaction problem

[pata320]

4PH3(g) ------- P4(g) + 6H2(g)
If, in a certain experiment, over a specific time period, 0.0063 mol PH3 is consumed in a 2.0 L container each second of reaction, what are the rates of production of P4 and H2 in this experiment.

Any idea about this problem? divide 2.0 to 1.0 and then .0063 to .00315 but from there I have no clue. Anybody have a clue? Thanks a bunch!

 

[sdekivit]

4PH3(g) ------- P4(g) + 6H2(g)
If, in a certain experiment, over a specific time period, 0.0063 mol PH3 is consumed in a 2.0 L container each second of reaction, what are the rates of production of P4 and H2 in this experiment.

the rate of consuming of PH3 is 0,0063 mol / s. Now use stoichiometry to get to know the rates of formation of P4 and H2:

1. when 4 mol PH3 has reacted, then 1 mol P4 is formed --> How much P4 is formed per second when 0,0063 mol PH3 has reacted (per second) ?

2. when 4 mol PH3 has reacted, then 6 mol H2 is formed --> How much H2 is formed per second when 0,0063 mol PH3 has reacted (per second) ?

 

[Blueshift5]

The rate of production for P4 and H2 can be calculated using the balanced chemical equation and the given information. First, we need to convert the volume of the container from 2.0 L to 1.0 L to match the stoichiometric ratio in the equation. This gives us 0.00315 mol PH3 consumed per second. 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. Therefore, the rate of production for P4 is 0.00315/4 = 0.0007875 mol/s and the rate of production for H2 is 0.00315/4*6 = 0.004725 mol/s. These rates represent the amount of P4 and H2 produced per second in the reaction.