The value of kp at 20°c for the reaction 2NO(g) + Cl2(g) 2NOCl(g) is 1.9x10^3 atm-1 calculate the value of KC at the same temperature.

Kp = Kc(RT)^delta n.

delta n = nproducts-nreactants

Answer

To calculate the value of Kc using the given value of Kp, we need to use the ideal gas law equation and the expression of Kc in terms of Kp.

The ideal gas law equation is given as:

PV = nRT

Where:
- P is the pressure of the gas in atm
- V is the volume of the gas in liters
- n is the number of moles of the gas
- R is the ideal gas constant (0.0821 atm·L/mol·K)
- T is the temperature in Kelvin

In the reaction 2NO(g) + Cl2(g) → 2NOCl(g), we can assume the reaction is taking place in a closed container at a fixed temperature (20°C).

Now, let's set up the equation for the reaction in terms of partial pressures:

Kp = (pNOCl)^2 / ((pNO)^2 * (pCl2))

Given:
Kp = 1.9 x 10^3 atm^(-1)

We need to find Kc, which is the equilibrium constant in terms of molar concentrations. The relationship between Kp and Kc is given by the following expression:

Kp = Kc * (RT)^(∆n)

Where:
∆n = (moles of products) - (moles of reactants)

In this case, ∆n = (2 - 2 - 1) = -1, as the reaction has one fewer mole of gas on the product side compared to the reactant side.

At the given temperature (20°C), we need to convert it to Kelvin:

20°C + 273.15 = 293.15 K

Now we can rearrange the equation to solve for Kc:

Kc = Kp / (RT)^(∆n)

Substituting the given values:

Kc = (1.9 x 10^3 atm^(-1)) / ((0.0821 atm·L/mol·K) * (293.15 K))^(-1)

Calculating the expression inside the parentheses:

Kc = (1.9 x 10^3) / (0.0821 * 293.15)^(-1)

Kc = (1.9 x 10^3) / (24.0582)^(-1)

Kc = (1.9 x 10^3) * 24.0582

Kc = 4.578 x 10^4

Therefore, the value of Kc at the same temperature is 4.578 x 10^4.