Question

If, in a certain experiment, over a specific time period, 5 mole of is consumed in a 50-L container each second of reaction, what are the rates of production of and in this experiment?

Answer 1

Check explanation

Step-by-step explanation:

Okay, let us start by looking for the rates of production of P4 gas.

Important things to know: From the question, 5 moles of the Reactant is used/consumed; the volume of the container is 50 litres per seconds. The equation of reaction is given below:

4PH3 (g) -------------------> P4 (g) + 6H2 (g). ------------------------------------(1).

So, from the reaction (1) above, four (4) moles of PH3 gives one mole of P4 and six(6) moles of Hydrogen.

Step one: calculate the number of moles of P4.

Therefore,

4 mole of PH3 = 1 mole of P4.

5 moles of PH3= x moles of P4.

Hence, x moles of P4= 5 moles of PH3 × 1 mole of P4÷ 4 moles of PH3.

So, we have x= 1.25 moles of P4.

Step two:

Determine the rate of formation of P4.

The rate of Reaction is the Increament in products concentration. Therefore, in 50 litres container;

(1.25 mole of P4÷ 50 litres per seconds)= 0.025 Mole per litre. Second is produced.

We will have to follow the same two steps for the determination of the rate of production of H2.

Step one: (5 moles of PH3 × 6 mole of H2 / 4 moles of PH3). = 7.5 moles of Hydrogen.

Step two: 7.5 moles of Hydrogen/ 50 litres per seconds.

==> 0.15 moles/litres. Seconds of H2..