#371 Numerical problems 1. At 721C, pº, after a pure Ha gas

Numerical problems 1. At 721C, pº, after a pure Ha gas - Chemistry

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II. Numerical problems 1. At 721C, pº, after a pure Ha gas slowly passes through excessive CoO(s), part of the oxide is reduced into Co(s). The flowing-out equilibrium gas contains 2.5% (volume fraction) Hz. At the same temperature, if we use CO gas to reduce Coo(s), the flowing-out equilibrium gas contains 1.92% (volume fraction ) CO. Then, if the same mole number of CO and water vapor are mixed up at 7210, and they undergo chemical reaction after an appropriate catalyst is added, how much is the equilibrium conversion percentage?

2. In the gas phase reaction 2A + B = 3C + 2D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and allowed to come to equilibrium at 25 C, the resulting mixture contained 0.90 mol C at a total pressure of 1.00 bar. Calculate (a) the mole fractions of each species at equilibrium, (b) K., (c) K, and {d} 4,GⓇ.

3. The dissociation vapour pressure of NH4Cl at 427°C is 608 kPa but at 459°C it has risen to 1115 kPa. Calculate (a) the equilibrium constant, (b) the standard reaction Gibbs energy, (e) the standard enthalpy, and (d) the standard entropy of dissociation, all at 427°C. Assume that the vapour behaves as a perfect gas and that AHⓇ and AS® are independent of temperature in the range given.
2. In the gas phase reaction 2A + B = 3C + 2D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and allowed to come to equilibrium at 25°C, the resulting mixture contained 0.90 mol C at a total pressure of 1.00 bar. Calculate (a) the mole fractions of each species at equilibrium, (b) Kx, (e) K, and {d} 4,6°.

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SOLN 1

The complete Balanced reaction is written as

CoO(s) + H2(g) = Co(s) + H2O(g)

Given: 1 mole of H2.

Since at equilibrium, the Volume fraction which is also called as mole fraction of H2 is 2.5%. therefore rest will be H2O which has mole fraction equal to 97.5%.

Equilibrium constant K1= [H2O] /[H2] = 0.975/0.025 = 38.88

From the above reaction, Co(s)+ H2O---àCoO+H2(g) (1), K1’ = 1/38.88= 0.026

CoO+CO(g) ---àCO2(g)+ Co (2)

Given: 1 mole of CO

Since volume fraction of CO is 1.92%. hence remaining will be CO2 which has mole fraction is =100-1.92= 98.08%

Equilibrium constant, K2= 0.9808/0.0192 = 51.08

combining reaction 1 and 2

H2O+CO----àCO2+H2, K = K1’*K2= 51.08/0.026 = 1.33

assuming 1 mole of H2O which has to be mixed with 1 mole of CO

Let assume that x= percentage decomposition of H2O

Therefore at equilibrium, [CO] = [H2O] =1-x and [CO2] =[H2] =x

Therefore x2/(1-x)2= 1.33 or x/(1-x)= 1.33 or x= 1.33-1.33x

2.33x= 1.33 or x= 1.33/2.33 =.578