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1.5 mol of PCl(5) are heated at constant...

`1.5 mol` of `PCl_(5)` are heated at constant temperature in a closed vessel of `4 L` capacity. At the equilibrium point, `PCl_(5)` is `35%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant.

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To calculate the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \), we will follow these steps: ### Step 1: Write the balanced chemical equation The dissociation of \( PCl_5 \) can be represented as: \[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \] ### Step 2: Determine the initial moles and changes at equilibrium We start with \( 1.5 \) moles of \( PCl_5 \) in a \( 4 \, L \) vessel. At equilibrium, \( 35\% \) of \( PCl_5 \) is dissociated. - Initial moles of \( PCl_5 \) = \( 1.5 \, mol \) - Moles of \( PCl_5 \) dissociated = \( 0.35 \times 1.5 = 0.525 \, mol \) - Moles of \( PCl_3 \) formed = \( 0.525 \, mol \) - Moles of \( Cl_2 \) formed = \( 0.525 \, mol \) ### Step 3: Calculate moles at equilibrium - Moles of \( PCl_5 \) at equilibrium = \( 1.5 - 0.525 = 0.975 \, mol \) - Moles of \( PCl_3 \) at equilibrium = \( 0.525 \, mol \) - Moles of \( Cl_2 \) at equilibrium = \( 0.525 \, mol \) ### Step 4: Calculate concentrations at equilibrium Concentration is calculated using the formula: \[ \text{Concentration} = \frac{\text{Moles}}{\text{Volume (L)}} \] - Concentration of \( PCl_5 \): \[ [C_{PCl_5}] = \frac{0.975 \, mol}{4 \, L} = 0.24375 \, M \] - Concentration of \( PCl_3 \): \[ [C_{PCl_3}] = \frac{0.525 \, mol}{4 \, L} = 0.13125 \, M \] - Concentration of \( Cl_2 \): \[ [C_{Cl_2}] = \frac{0.525 \, mol}{4 \, L} = 0.13125 \, M \] ### Step 5: Write the expression for the equilibrium constant \( K_c \) The expression for the equilibrium constant \( K_c \) for the reaction is: \[ K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} \] ### Step 6: Substitute the concentrations into the \( K_c \) expression \[ K_c = \frac{(0.13125)(0.13125)}{0.24375} \] ### Step 7: Calculate \( K_c \) Calculating the value: \[ K_c = \frac{0.017265625}{0.24375} \approx 0.0708 \] ### Final Answer The equilibrium constant \( K_c \) is approximately \( 0.0708 \, M^{-1} \). ---

To calculate the equilibrium constant \( K_c \) for the dissociation of \( PCl_5 \) into \( PCl_3 \) and \( Cl_2 \), we will follow these steps: ### Step 1: Write the balanced chemical equation The dissociation of \( PCl_5 \) can be represented as: \[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \] ...
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1*5 moles of PCl_(5) are heated at constant temperature in a closed vessel of 4 litre capacity. At the equilibrium point, PCl_(5)" is "35 % " dissociated into " PCl_(3) and Cl_(2). Calculate the equilibrium constant.

2 "mole" of PCl_(5) were heated in a closed vessel of 2 litre capacity. At equilibrium 40% of PCl_(5) dissociated into PCl_(3) and Cl_(2) . The value of the equilibrium constant is:

Knowledge Check

  • Two moles of PCl_(5) were heated in a closed vessel of 2 L. At equilibrium 40% of PCl_(5) is dissociated into PCl_(3) and Cl_(2) . The value of equilibrium constant is

    A
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    B
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    D
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    A
    `0.266`
    B
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    D
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  • 2 g molecule of PCl_5 are heated in a closed vessel of two litre capacity. When the equilibrium is attained , PCl_5 is 40% dissociated into PCl_3 and Cl_2 . The equilibrium constant is

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    B
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    C
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    D
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