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Equilibrium concentrations of A and B in...

Equilibrium concentrations of A and B involved in following equilibrium are 0.01 M and 0.02 M respectively: `[A(g)implies B(g)]` If 0.01 M of A is added at equilibrium, then at new equilibrium concentration of B will be

A

`6.67` x `10^-2` M

B

`2.67` x `10^-2` M

C

`1.33` x `10^-2` M

D

`1.33` x `10^-1` M

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To solve the problem step by step, we will follow these steps: ### Step 1: Understand the equilibrium concentrations We are given the equilibrium concentrations of A and B: - \([A]_{eq} = 0.01 \, M\) - \([B]_{eq} = 0.02 \, M\) ### Step 2: Write the equilibrium expression The equilibrium expression for the reaction \( A(g) \rightleftharpoons B(g) \) is given by: \[ K_c = \frac{[B]}{[A]} \] Substituting the equilibrium concentrations: \[ K_c = \frac{0.02}{0.01} = 2 \] ### Step 3: Add 0.01 M of A to the system When we add 0.01 M of A to the system, the new concentration of A becomes: \[ [A]_{new} = 0.01 + 0.01 = 0.02 \, M \] The concentration of B remains the same initially at 0.02 M. ### Step 4: Set up the reaction shift According to Le Chatelier's principle, adding more reactant (A) will shift the equilibrium to the right (towards the products, B). Let \( x \) be the amount of A that reacts to form B. At the new equilibrium: - \([A] = 0.02 - x\) - \([B] = 0.02 + x\) ### Step 5: Write the new equilibrium expression The equilibrium expression remains the same: \[ K_c = \frac{[B]}{[A]} = 2 \] Substituting the new concentrations: \[ 2 = \frac{0.02 + x}{0.02 - x} \] ### Step 6: Solve for \( x \) Cross-multiplying gives: \[ 2(0.02 - x) = 0.02 + x \] Expanding and rearranging: \[ 0.04 - 2x = 0.02 + x \] \[ 0.04 - 0.02 = 2x + x \] \[ 0.02 = 3x \] \[ x = \frac{0.02}{3} = 0.00667 \, M \] ### Step 7: Calculate the new concentration of B Now, substituting \( x \) back to find the new concentration of B: \[ [B]_{new} = 0.02 + x = 0.02 + 0.00667 = 0.02667 \, M \] ### Step 8: Final answer The new concentration of B at equilibrium is approximately: \[ [B]_{new} \approx 0.02667 \, M \text{ or } 0.0267 \, M \]
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AAKASH INSTITUTE-TEST 2-EXERCISE
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