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Determinre the value of equilibrium cons...

Determinre the value of equilibrium constant `(K_(C))` for the reaction
`A_(2)(g)+B_(2)(g)hArr2AB(g)`
if 10 moles of `A_(2)` ,15 moles of`B_(2)` and 5 moles of AB are placed in a 2 litre vessel and allowed to come to equilibrium . The final concentration of AB is 7.5 M:

A

`4.5`

B

`1.5`

C

`0.6`

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To determine the equilibrium constant \( K_c \) for the reaction \[ A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \] given the initial moles and the final concentration of \( AB \), we will follow these steps: ### Step 1: Write the Initial Conditions Initially, we have: - Moles of \( A_2 = 10 \) - Moles of \( B_2 = 15 \) - Moles of \( AB = 5 \) The volume of the vessel is 2 liters. ### Step 2: Calculate Initial Concentrations The initial concentrations can be calculated using the formula: \[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \] Thus, the initial concentrations are: - \( [A_2] = \frac{10}{2} = 5 \, M \) - \( [B_2] = \frac{15}{2} = 7.5 \, M \) - \( [AB] = \frac{5}{2} = 2.5 \, M \) ### Step 3: Set Up the Change in Concentrations Let \( x \) be the amount of \( A_2 \) and \( B_2 \) that react at equilibrium. The changes in concentration will be: - \( [A_2] \) decreases by \( x \) - \( [B_2] \) decreases by \( x \) - \( [AB] \) increases by \( 2x \) (since 1 mole of \( A_2 \) produces 2 moles of \( AB \)) At equilibrium, the concentrations will be: - \( [A_2] = 5 - x \) - \( [B_2] = 7.5 - x \) - \( [AB] = 2.5 + 2x \) ### Step 4: Use Given Final Concentration of \( AB \) We know the final concentration of \( AB \) is 7.5 M. Therefore, we can set up the equation: \[ 2.5 + 2x = 7.5 \] ### Step 5: Solve for \( x \) Rearranging the equation gives: \[ 2x = 7.5 - 2.5 \] \[ 2x = 5 \] \[ x = 2.5 \] ### Step 6: Calculate Equilibrium Concentrations Now we can find the equilibrium concentrations: - \( [A_2] = 5 - 2.5 = 2.5 \, M \) - \( [B_2] = 7.5 - 2.5 = 5 \, M \) - \( [AB] = 7.5 \, M \) (as given) ### Step 7: Write the Expression for \( K_c \) The expression for the equilibrium constant \( K_c \) for the reaction is: \[ K_c = \frac{[AB]^2}{[A_2][B_2]} \] ### Step 8: Substitute the Equilibrium Concentrations Substituting the equilibrium concentrations into the expression: \[ K_c = \frac{(7.5)^2}{(2.5)(5)} \] ### Step 9: Calculate \( K_c \) Calculating the values: \[ K_c = \frac{56.25}{12.5} = 4.5 \] ### Final Answer Thus, the value of the equilibrium constant \( K_c \) is: \[ \boxed{4.5} \]
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