Question

A flask is charged with 1.350 atm of N2O4(g) and 1.00 atm of NO2(g) at 25 ∘C, and the following equilibrium is achieved:

N2O4(g)⇌2NO2(g)

After equilibrium is reached, the partial pressure of NO2 is 0.517 atm.

Part A:

What is the partial pressure of N2O4 at equilibrium?

Express the pressure to three significant figures and include the appropriate units

Part B:

Calculate the value of Kp for the reaction.

Express your answer to three significant figures.

Part C:

Calculate the value of Kc for the reaction.

Express your answer to three significant figures.

Answer 1

Sol:-

(a). ICE table of the given reaction is :

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

Initial (I)......................1.350 atm.................................................................1.00 atm

Change (C).................+α..............................................................................-2α

Equilibrium (E)..........(1.350+α) atm...........................................................(1.00-2α)

Here, α = Amount dissociated per mole

Now,

Given, equilibrium partial pressure of NO₂ = 0.517 atm

So,

(1.00-2α) = 0.517 atm

So,

α = 0.2415

So,

Equilibrium partial pressure of N₂O₄ = (1.350+α) atm = (1.350+0.2415) atm = 1.5915 atm

Hence, Equilibrium partial pressure of N₂O₄ = 1.59 atm

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(b).

Expression of Pressure equilibrium constant (K_p) is : Which is equal to the ratio of product of the partial pressure of products to the product of the partial pressure of reactants raise to power of stoichiometric coefficient at equilibrium stage of the reaction.

K_p = P_NO2² / P_N2O4

K_p = (0.517 atm)² / (1.59 atm)

K_p = 0.168 atm

Hence, value of K_p for the reaction = 0.168 atm

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(c).

Relationship between K_p and K_c is :

K_p = K_c(RT)^Δn_g

K_c = K_p (RT)^-Δn_g

K_c = (0.168 atm).(0.0821 Latm K^-1mol^-1 x 298 K)^-(2-1)

K_c = 6.87 x 10^-3 M

Hence,K_c = 6.87 x 10^-3 M