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.

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

According to the question,

→

 1.500       1.00

    +a         -2a

1.500+a   1.00-2a  

At equilibrium,

→

a = 0.2405

→

the,

→ The Kp of the reaction will be:

=

=

=

Now,

The change in moles,

Temperature,

Molar gas constant,

then,

→ The Kc for the reaction will be:

=

=

=

Thus the answer above is correct.

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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