Question

In: Chemistry

#1. In the reaction: C2H4(g) + H2O(g) <---> C2H5OH(g) DeltaH = -47.8 kJ Kc = 9.00*103...

#1. In the reaction: C₂H₄(g) + H₂O(g) <---> C₂H₅OH(g) DeltaH = -47.8 kJ K_c = 9.00*10³ @ 600. K

a. At equilibrium, P_C2H5OH = 200. atm & P_H2O = 400. atm. Calculate P_C2H4.

b. Which set of conditions results in the highest yield of product.

Pressure high/low __________ Temperature high/low ________

Expert Solution

Answer – We are given, reaction –

C₂H₄(g) + H₂O(g) <---> C₂H₅OH(g) ∆H = -47.8 kJ K_c = 9.00*10³ T= 600. K

At equilibrium, P_C2H5OH = 200. atm & P_H2O = 400. Atm

First we need to calculate the Kp form the Kc

Kp = Kc (RT)^∆n

We know,

∆n = sum of the moles of the product – reactant

= 1 – 2 = -1

Kp = 9.00*10³ *(0.0821 *600)^-1

=182.7

Now we know

Kp = P_H2O * P_C2H4 / P_C2H5OH

182.7 = 400.0 atm * P_C2H4 /200.0 atm

182.7*200 = 400* P_C2H4 /

36540 = 400* P_C2H4

P_C2H4 = 91.35 atm

b) We are given the negative enthalpy of reaction and we calculated the Kp of the reaction. We know when negative enthalpy then it is exothermic reaction, so as we increase the temp there is decrease the product. We also know when we increase the pressure the equilibrium will shift towards the fewer moles of side, so at the conditions pressure is high and temperature low gives highest yield of product.

queen_honey_blossom answered 2 years ago

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