Question

Question 2  

1. Nitrogen dioxide, ??₂, exists in equilibrium with dinitrogen tetroxide, ?₂?₄, according to the reaction

equation: ?₂?₄(g) ⇌ 2 ??₂(g)

When 1.566 ? of ?₂?₄ is present in a 1.00 ??³ vessel at 25.0 °C, the pressure is 0.597 ???.

(a) Calculate the number of moles of ?₂?₄(g) before any dissociation.                                         [4]

(b) Write the total number of moles of the gas mixture at equilibrium in terms of the degree of

dissociation, α.                                                                                                                                  [4]

(c) Calculate the total equilibrium amount, ?, of the gas mixture.                                                  [4]

(d) Calculate the extent of dissociation, α, under the above mentioned conditions.                        [4]

Give the final answer to two decimal places!

(e) Calculate the equilibrium constant ?.                                                                                        [4]

Answer 1

Ans :-

(a). Given mass of N₂O₄ gas = 1.566 g

Moles of N₂O₄ gas = Mass in g /Gram molar mass

= 1.566 g / 92.011 g/mol

= 0.01702 mol

Therefore, number of moles of N₂O₄ gas before dissociation = 0.01702 mol

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(b). ICE table is :

..........................N₂O₄ (g) <--------------------> 2NO₂ (g)

Initial .................0.01702 mol.........................0.0 mol

Change.................+α........................................-2α

Equilibrium .......(0.01702+α) mol.....................-2α mol

Therefore, Total number of moles of gases at equilibrium = 0.01702 + α - 2α = (0.01702 - α) mol

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(c). Number of moles of N₂O₄ (g) (n) = PV/RT

= (0.58919319 atm).(1.00 L) / (0.0821 L atm K^-1mol^-1).(298 K)

= 0.02408 mol

Therefore, 0.01702+α = 0.02408

α = 0.00706

So, Total equilibrium amount, ?, of the gas mixture = (0.01702 - 0.00706) mol = 0.00996 mol

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(d). The extent of dissociation α, under the above mentioned conditions = α = 0.00706

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(e). Expression of Equilibrium constant i.e. K (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K = [NO₂ ]²/[N₂O₄]

K = (2α)²/(0.01702-α)

K = 4α²/(0.01702-α)

K = 4 x (0.00706)² / (0.01702 - 0.00706)

K = 4 x (0.00706)² / (0.00996)

K = 2.0 x 10^-2

Therefore, equilibrium constant = K = 2.0 x 10^-2