The equilibrium constant for the chemical equation N2(g) + 3H2(g) ⇌ 2NH3(g) and Kp=0.174 at 243°C. Calculate the value of Kc for the reaction at 243°C.

The Kc of this reaction is 311.97

Explanation:

Step 1: Data given

Kp = 0.174

Temperature = 243 °C

Step 2: The balanced equation

N2(g) + 3H2(g) ⇌ 2NH3(g)

Step 3: Calculate Kc

Kp = Kc *(RT)^Δn

⇒ with Kp = 0.174

⇒ with Kc = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 Latm/Kmol

⇒ with T = the temperature = 243 °C = 516 K

⇒ with Δn = number of moles products - moles reactants  2 – (1 + 3) = -2

0.174 = Kc (0.08206*516)^-2

Kc = 311.97

The Kc of this reaction is 311.97

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :