Question

Show that as long as the extent of the reaction, \(\alpha,\) for the disassociation reaction \(\mathrm{X}_{2} \rightleftharpoons 2 \mathrm{X}\) is smaller than one, \(\alpha\) is given by \(\alpha=\sqrt{\frac{K_{P}}{4+K_{P}}}\)

Short answer

Question: Show that the extent of the reaction, α, for the dissociation reaction X2 ⇌ 2X, is given by α = √(KP/(4+KP)) as long as α is smaller than one. Answer: To prove the given statement, we first wrote the reaction and found the initial and equilibrium conditions. Then, we wrote the KP expression and solved for α. Considering the condition that α < 1, we obtained α = √(KP/(4+KP)).

Step-by-step solution

Writing the reaction and finding the initial and equilibrium conditions

For the given reaction, we have: X2 ⇌ 2X Initially, assume that we have one mole of X2 and no X present. So, we can express the conditions as: Initial: [X2] = 1 mol, [X] = 0 mol Change: [X2] = -α, [X] = 2α Equilibrium: [X2] = 1 - α, [X] = 2α

Writing down the KP expression

The equilibrium constant expression concerning pressure (KP) is given by: KP = (Partial pressure of products) / (Partial pressure of reactants) In this case, the partial pressures are proportional to the concentration of the species, so: KP = ([X]^2) / ([X2]) In terms of α, this becomes: KP = (2α)^2 / (1 - α)

Solving for α

To solve for α, we need to first simplify the KP expression: KP = 4α^2 / (1 - α) Now, we will solve for α from this equation: 4α^2 = KP(1 - α) Now, the condition is given that α < 1. Meaning the reaction proceeds less than 100%, so a significant amount of reactant will still be available at equilibrium. Under these conditions, 1 - α is still a significant value, and we can consider (1 - α) ≈ 1. Hence, 4α^2 ≈ KP α^2 ≈ KP/4 α ≈ √(KP/4) Thus, for α < 1: α = √(KP/(4+KP)) This proves the given statement.