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For the reaction 2A(g)hArrB(g)+3C(g), at...

For the reaction `2A(g)hArrB(g)+3C(g),` at a given temperature ,`K_(c)=16.` What must be the volume of the flask , if a mixture of 2 mole rach of A,B and C exist in equilibrium ?

A

`(1)/(4)`

B

`(1)/(2)`

C

1

D

none of these

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: \[ 2A(g) \rightleftharpoons B(g) + 3C(g) \] ### Step 2: Define the equilibrium constant expression The equilibrium constant \( K_c \) for the reaction is defined as: \[ K_c = \frac{[B]^{b} \cdot [C]^{c}}{[A]^{a}} \] where \( a, b, c \) are the stoichiometric coefficients from the balanced equation. For our reaction: \[ K_c = \frac{[B] \cdot [C]^3}{[A]^2} \] ### Step 3: Set up the concentrations We are given that there are 2 moles of each species (A, B, and C) at equilibrium. Let the volume of the flask be \( V \) liters. Therefore, the concentrations of each species will be: - Concentration of \( A \): \([A] = \frac{2}{V}\) - Concentration of \( B \): \([B] = \frac{2}{V}\) - Concentration of \( C \): \([C] = \frac{2}{V}\) ### Step 4: Substitute concentrations into the equilibrium expression Now substituting the concentrations into the \( K_c \) expression: \[ K_c = \frac{\left(\frac{2}{V}\right) \cdot \left(\frac{2}{V}\right)^3}{\left(\frac{2}{V}\right)^2} \] ### Step 5: Simplify the equation Substituting and simplifying: \[ K_c = \frac{\left(\frac{2}{V}\right) \cdot \left(\frac{8}{V^3}\right)}{\left(\frac{4}{V^2}\right)} = \frac{\frac{16}{V^4}}{\frac{4}{V^2}} = \frac{16}{4} \cdot \frac{1}{V^2} = \frac{4}{V^2} \] ### Step 6: Set \( K_c \) equal to the given value We know that \( K_c = 16 \), so we set up the equation: \[ \frac{4}{V^2} = 16 \] ### Step 7: Solve for \( V^2 \) Cross-multiplying gives: \[ 4 = 16V^2 \] \[ V^2 = \frac{4}{16} = \frac{1}{4} \] ### Step 8: Solve for \( V \) Taking the square root of both sides: \[ V = \sqrt{\frac{1}{4}} = \frac{1}{2} \] ### Final Answer The volume of the flask must be: \[ \boxed{\frac{1}{2} \text{ liters}} \]
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