When hydrogen is removed from an alkane, an alkene is formed - Edexcel - GCSE Chemistry - Question 10 - 2022 - Paper 1

To maximize the yield of propene from propane, the manufacturer should consider the following conditions:

1. Increase Temperature: Since the forward reaction is endothermic, increasing the temperature shifts the equilibrium position toward the production of propene and hydrogen, thus favoring the formation of products.

2. Decrease Pressure: Lowering the pressure can favor the production of gases with a larger number of moles. In this case, since one mole of propane converts to two moles of products (propene and hydrogen), reducing pressure will help to shift the equilibrium toward the side with more gas moles.

3. Use of Catalyst: Incorporating a catalyst can increase the rate of reaction without affecting the equilibrium position. A catalyst would lower the activation energy required for the reaction to proceed, thereby speeding up the production rate of propene.

In the dehydrogenation reaction:

$C₃H₈(g) ⇌ C₃H₆(g) + H₂(g)$

From the equation, 1 mole of propane produces 1 mole of propene. Therefore, if 300 dm³ of propane is completely converted, the maximum volume of propene formed will also be 300 dm³.

From the reaction:

$C₃H₈(g) ⇌ C₃H₆(g) + H₂(g)$

Here, 1 mole of propane produces 1 mole of hydrogen. Given that 900 dm³ of propane is reacted:

• The amount of propane (in moles) can be calculated using the molar volume (24 dm³/mol):

  $n = \frac{900 \text{ dm}^3}{24 \text{ dm}^3/\text{mol}} = 37.5 \text{ moles}$

• This means that 37.5 moles of hydrogen are produced.

• The mass of hydrogen can be calculated using its molar mass (1 g/mol):

  $\text{Mass} = n \times \text{molar mass} = 37.5 \text{ moles} \times 1 \text{ g/mol} = 37.5 ext{ g}$

• Converting grams to kilograms:

  $37.5 ext{ g} = 0.0375 ext{ kg}$

Thus, the maximum mass of hydrogen formed is 0.0375 kg.