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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

PCl_ 5 vapour decomposes on heating according to the reaction: PCl...

{\text{PCl}_{5} vapour decomposes on heating according to the reaction:

PCl5(g)⇌PCl3(g)+Cl2(g){PCl}_{5} \left(\text{g}\right)⇌ {PCl}_{3} \left(\text{g}\right) + {Cl}_{2}\left(\text{g}\right)

The density of a sample of a partially dissociated PCl{\text{PCl}}_{5}at1.0atmand500Kwasfound4.8g/L.CalculatethedegreeofdissociationandΔat \text{1}\text{.0} atm and \text{500}\text{K} was found \text{4}\text{.8}\text{g/L}\text{.} Calculate the degree of dissociation and \Delta G^{∘}forthereactionat500K.<br/>(Given:R=0.082LatmKfor the reaction at \text{500}\text{K}\text{.}<br /> (Given: \text{R} =0.082{\text{LatmK}}^{- 1} mol{\text{mol}}^{- 1}, R=8.314JK\text{R} =8.314{\text{JK}}^{- 1} mol{\text{mol}}^{- 1},lnx=2.303log⁡10x)ln\text{x} =2.303\log_{10} \text{x} )

Correct answer

23.40

Step-by-step solution

Mavg=ρRTP=4.8×0.082×5001=196.8{M}_{avg} = \dfrac{ρ \text{RT}}{\text{P}} = \dfrac{4 . 8 \times 0 . 0 8 2 \times 5 0 0}{1} = 1 9 6 . 8

{PCl5(g)⇌PCl3(g)+Cl2(g)1−ααα(Total moles=1+α)\begin{cases}{PCl}_{5} \left(\text{g}\right) & ⇌ & {PCl}_{3} \left(\text{g}\right) & + & {Cl}_{2} \left(\text{g}\right) & \\ 1 - \alpha & & \alpha & & \alpha & \left(\text{Total moles} = 1 + \alpha\right)\end{cases}

Mavg=MPCl51+α=196.8=208.51+α{M}_{avg} = \dfrac{{M}_{\text{PCl} 5}}{1 + \alpha} = 1 9 6 . 8 = \dfrac{2 0 8 . 5}{1 + \alpha}

∴α=0.06∴ \alpha = 0 . 0 6

KP=α21−α2P=(0.06)21−(0.06)2=3.6×10−3{K}_{P} = \dfrac{{α}^{2}}{1 - {α}^{2}} \text{P} = \dfrac{{\left(0.06\right)}^{2}}{1 - {\left(0.06\right)}^{2}} = 3.6 \times {10}^{- 3}

ΔG∘=−RTlnK\Delta G^{∘} = - \text{RT} \text{ln} \text{K}

& & =−8.314×500×2.303log(3.6×10−3)= - 8 . 3 1 4 \times 5 0 0 \times 2 . 303 \text{log} \left(3 . 6 \times1 {0}^{- 3}\right)

& & =23.4kJ= 23.4\text{kJ}
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