Question

Calculate the number of moles of Cl2 produced at equilibrium when one mole of PCl5 is heated at

250 in vessel having a capacity of 10dm3

(Kc=0.041) (6)

PCl5↔ PCl3 + Cl2​

Answer 1

Answer:

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Answer 2

Given:

One mole of PCl₅ is heated at 250⁰C in a vessel of capacity of 10dm³.          Kc=0.041

To find:

Number of moles of Cl₂ produced at equilibrium.

Solution:

The required chemical reaction is-

   PCl₅ (g)    →    PCl₃ (g)  +   Cl₂ (g)

at t=0, concentration of PCl₅ , ${{\text{C}}} = \left {\frac{{n }}{{V }}}$

                                                 ${{\text{C}}} = \left {\frac{{1 }}{{10 }}}$

                                                  C= 0.1 moles/dm³

initially, concentration of PCl₅ is 0.1moles/dm³ whereas conc. of PCl₃ and Cl₂ is 0.

suppose, x moles/dm³ of PCl₃ and Cl₂ is formed at equilibrium then conc. of PCl₅ at equilibrium is (0.1-x) moles/dm³

So, rate constant at equilibrium is

${{\text{Kc}}} = \left {\frac{{[PCl_3] \cdot [Cl_2] }}{{[PCl_5] }}}$

${{\text{0.041}}} = \left {\frac{{x \cdot x }}{{(0.1-x) }}}$

by solving the above equation, value of x will be

x = 0.04 (approximately)

thus, concentration of Cl₂(x) = 0.04 moles/dm³

number of moles of Cl₂ = concentration of Cl₂ × volume

                                       = 0.04 moles/dm³ × 10 dm³

                                        = 0.4 moles

So, number of moles of Cl₂ is 0.4moles.