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In the reaction X((g))+ Y((g)) hArr 2Z((...

In the reaction `X_((g))+ Y_((g)) hArr 2Z_((g))`, 2 moles of X, I mole of Y and I mole of Z are placed in a 10 lit vessel and allowed to reach equilibrium. If final concentration of Z is 0.2 M, then `K_(c)`. for the given reaction is

A

1.6

B

`(80)/(3)`

C

`(16)/(3)`

D

`(3)/(16)`

Text Solution

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The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Write the balanced chemical equation The reaction is given as: \[ X_{(g)} + Y_{(g)} \rightleftharpoons 2Z_{(g)} \] ### Step 2: Determine the initial moles of each substance We have: - Initial moles of \(X = 2\) moles - Initial moles of \(Y = 1\) mole - Initial moles of \(Z = 1\) mole ### Step 3: Calculate the initial concentrations The volume of the vessel is 10 L. Therefore, the initial concentrations are: - \([X]_0 = \frac{2 \text{ moles}}{10 \text{ L}} = 0.2 \, M\) - \([Y]_0 = \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \, M\) - \([Z]_0 = \frac{1 \text{ mole}}{10 \text{ L}} = 0.1 \, M\) ### Step 4: Set up the change in concentration at equilibrium Let \(a\) be the amount of \(X\) that reacts at equilibrium. Then, the changes in concentration will be: - For \(X\): \(0.2 - a\) - For \(Y\): \(0.1 - a\) - For \(Z\): \(0.1 + 2a\) ### Step 5: Use the information about the equilibrium concentration of \(Z\) We know that the final concentration of \(Z\) is \(0.2 \, M\): \[ 0.1 + 2a = 0.2 \] Solving for \(a\): \[ 2a = 0.2 - 0.1 = 0.1 \implies a = 0.05 \] ### Step 6: Substitute \(a\) back to find equilibrium concentrations Now we can find the equilibrium concentrations: - \([X]_{eq} = 0.2 - a = 0.2 - 0.05 = 0.15 \, M\) - \([Y]_{eq} = 0.1 - a = 0.1 - 0.05 = 0.05 \, M\) - \([Z]_{eq} = 0.1 + 2a = 0.1 + 0.1 = 0.2 \, M\) ### Step 7: Write the expression for \(K_c\) The equilibrium constant \(K_c\) for the reaction is given by: \[ K_c = \frac{[Z]^2}{[X][Y]} \] Substituting the equilibrium concentrations: \[ K_c = \frac{(0.2)^2}{(0.15)(0.05)} \] ### Step 8: Calculate \(K_c\) Calculating the values: \[ K_c = \frac{0.04}{0.0075} = \frac{40}{7.5} = \frac{80}{15} = \frac{16}{3} \] ### Final Answer Thus, the value of \(K_c\) for the given reaction is: \[ K_c = \frac{16}{3} \] ---
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