Question

Calculate the equilibrium constants Kp and Kc for the reaction, CO (8) + 1/2 02= co2
Given that the partial pressures at equilibrium in a vessel at 3000 K are Pco = 0.4 atm. Pco2 = 0.6 am
Po, = 0.2 atm​

Answer 1

Answer: B

CO(g)+12O2(g)⇔CO2(g)

Kp=[pCO2(g)][pCO(g)][pO2(g)]1/2

=0.60.4×(0.2)1/2=0.60.4×0.4472=3.356atm−1/2

Kp=Kc(RT)Δn  

Δn=1−(1+12)=−12

Kc=Kp(RT)Δn

=3.356(0.082×3000)−1/2

=3.356×(0.082×3000)1/2

=3.356×15.684=52.67

Explanation: