For the reaction 2 HCN(g) <--> H 2 (g) +C 2 N 2 (g) It is known that the equilib

Question

For the reaction
2 HCN(g) <--> H2(g) +C2N2(g)
It is known that the equilibrium constant at some temperatureis Kp=4.00 x 10-4. What is the equilibrium partialpressure for HCN if initially 3.00 moles of HCN is placed in 2.00 Lcontainer and no products are present?

2 HCN(g) <--> H2(g) +C2N2(g)
It is known that the equilibrium constant at some temperatureis Kp=4.00 x 10-4. What is the equilibrium partialpressure for HCN if initially 3.00 moles of HCN is placed in 2.00 Lcontainer and no products are present?

Explanation / Answer

P = (n/V)*RT = (3 mol/ 2 L)*(8.314 J/mol/K)*(298 K)*( 1atm/101325Pa) = 0.0366776018 atm                 2HCN(g)            H2(g) + C2N2(g) initial        3.668e-2 change    -2x       +x +x equil 3.668e-2 - 2x       +x       +x Kp = 4e-4 = [H2][C2N2]/[HCN]^2 = (x)*x)/(3.668e-2 -2x)2 Take square root x/(3.668e-2 -2x) = 2e-2 x/2e-2 = 3.668e-2 - 2x x = 0.000705338496 P_HCN = 3.668e-2 - 2x = 0.035269323 = 3.53 e-2 atm

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