Question

A flask is charged with 1.500atm of N2O4(g) and 1.00 atm NO2(g) at 25 degree C, and the following equilibrium is achieved: N2O4(g) 2NO2 After equilibrium is reached, the partial pressure of NO2 is 0.519atm. Calculate the value of Kp for the reaction. Calculate Kc for the reaction.

Answer 1

Answer:

Kp = 0.155

Kc = 0.006338

Explanation:

Step 1: Data given

Pressure N2O4 = 1.500 atm

Pressure NO2 = 1.00 atm

Temperature = 25.0 °C

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Initial pressure

pN2O4 = 1.500 atm

pNO2 = 1.00 atm

Step 3: Calculate the pressure at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

pNO2 = (1.00 - 2x) = 0.519 atm

x = 0.2405 atm

partial pressure N2O4 = 1.5 + 0.2405 = 1.7405 atm

Kp = (pNO2)² / (pN2O4)

Kp = (0.519)² / 1.7405 =0.155

Kp = Kc * RT(Δn)

⇒ with Kp = 0.155

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 25.0 °C = 298 K

⇒ with Δn = difference in moles between products and reactants = 2-1 = 1

0.155 = Kc * 0.08206 * 298 K * (1)

Kc = 0.006338

Answer 2

The Kp will be "0.155" and the Kc will be "0.00633".

According to the question,

→ $N_2O_4 \rightleftharpoons 2NO_2$

 1.500       1.00

    +a         -2a

1.500+a   1.00-2a  

At equilibrium,

→ $P(NO_2) = 1.00-2a = 0.519 \ atm$

a = 0.2405

→ $P(N_2O_4) = 1.500 +a =1.7405 \ atm$

the,

→ The Kp of the reaction will be:

= $\frac{P(NO_2)^2}{P(N_2O_4)}$

= $\frac{0.519^2}{1.7405}$

= $0.155$

Now,

The change in moles,

• $\Delta n = 2-1 =1$

Temperature,

• $T = 25^{\circ} C = 298.15 \ K$

Molar gas constant,

• $R = 0.08206 \ atm.L/mol.K$

then,

→ The Kc for the reaction will be:

= $K_p\times (RT)^{-\Delta n}$

= $0.1548\times (0.08206\times 298.15)^{-1}$

= $0.00633$

Thus the answer above is correct.

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