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In the reaction X(g)+Y(g)hArr2Z(g),2 mol...

In the reaction `X(g)+Y(g)hArr2Z(g),2` mole of X,1 mole of Y and 1 mole of Z are placed in a 10 litre vessel and allowed to reach equilibrium .If final concentration of Z is 0.2 M , then `K_(c)`for the given reaction is :

A

`1.60`

B

`(80)/(3)`

C

`(16)/(3)`

D

none of these

Text Solution

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To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction: \[ X(g) + Y(g) \rightleftharpoons 2Z(g) \] Given: - Initial moles of \( X = 2 \) - Initial moles of \( Y = 1 \) - Initial moles of \( Z = 1 \) - Volume of the vessel = 10 L - Final concentration of \( Z = 0.2 \, M \) ### Step 1: Calculate Initial Concentrations First, we calculate the initial concentrations of \( X \), \( Y \), and \( Z \): \[ \text{Concentration of } X = \frac{2 \, \text{moles}}{10 \, \text{L}} = 0.2 \, M \] \[ \text{Concentration of } Y = \frac{1 \, \text{mole}}{10 \, \text{L}} = 0.1 \, M \] \[ \text{Concentration of } Z = \frac{1 \, \text{mole}}{10 \, \text{L}} = 0.1 \, M \] ### Step 2: Set Up the Change in Concentration Let \( a \) be the change in moles of \( X \) and \( Y \) that react to form \( Z \). According to the stoichiometry of the reaction: - For every 1 mole of \( X \) and \( Y \) that reacts, 2 moles of \( Z \) are produced. - Therefore, at equilibrium: - Moles of \( X \) = \( 2 - a \) - Moles of \( Y \) = \( 1 - a \) - Moles of \( Z \) = \( 1 + 2a \) ### Step 3: Use the Final Concentration of \( Z \) We know the final concentration of \( Z \) is \( 0.2 \, M \): \[ \text{Concentration of } Z = \frac{1 + 2a}{10} = 0.2 \] Multiplying both sides by 10: \[ 1 + 2a = 2 \] Solving for \( a \): \[ 2a = 2 - 1 \implies 2a = 1 \implies a = \frac{1}{2} \] ### Step 4: Calculate Equilibrium Concentrations Now we can calculate the equilibrium concentrations: - Concentration of \( X \): \[ \frac{2 - a}{10} = \frac{2 - \frac{1}{2}}{10} = \frac{1.5}{10} = 0.15 \, M \] - Concentration of \( Y \): \[ \frac{1 - a}{10} = \frac{1 - \frac{1}{2}}{10} = \frac{0.5}{10} = 0.05 \, M \] - Concentration of \( Z \): \[ \frac{1 + 2a}{10} = \frac{1 + 1}{10} = \frac{2}{10} = 0.2 \, M \] ### Step 5: Calculate \( K_c \) The equilibrium constant \( K_c \) is given by: \[ K_c = \frac{[Z]^2}{[X][Y]} \] Substituting the equilibrium concentrations: \[ K_c = \frac{(0.2)^2}{(0.15)(0.05)} = \frac{0.04}{0.0075} \] Calculating \( K_c \): \[ K_c = \frac{0.04}{0.0075} = \frac{4}{0.75} = \frac{4 \times 4}{3} = \frac{16}{3} \] ### Final Answer Thus, the equilibrium constant \( K_c \) for the reaction is: \[ K_c = \frac{16}{3} \] ---
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