What would be the pressure of ethanol after equilibrium is reached the second time? Round to 2 decimal places

Question

C2H4(g)+H2O=CH3CH2OH(g)

Kc=[CH3CH2OH]/[C2H4][H2O]

A chemical engineer studying this fills 2.0L w/ 2.7atm of ethylene gas & 2.3atm of water vapor. When the mixture has come to equilibrium he determines that it contains 2.03 atm of ethylene gas and 1.63 atm of water vapor. The engineer then adds another 0.90 atm of ethylene and allows the mixture to come to equilibrium again.

*The topic is titled: Calculating an equilibrium composition after a prior equilibrium determines K

Answer 1

1st equilibrium: solve for K_P using K expression: P_Alc/(P_EthP_WV) You find P_Alc at eq from P_Eth,0 - P_Eth,eq = 2.7 - 2.03 = .67 atm and P_W and P_Eth are given for 1st equilibrium. (I'm using Alc for ethanol)

Now you know K for these conditions. Do an ICE calculation to determine final equilibrium concentrations in partial pressures. Since volume is constant we can use partial pressures as surrogates for moles of a particular species. Let x be amount of Alcohol formed when equilibrium system responds to addition of .9 atm of ethylene

Eth Water Alc

Initial 2.03 +.9 1.63 .67

Change -x -x +x

Equil 2.93-x 1.63-x .67+x

Solve quadratic equation for x: K(from part 1) = (.67+x)/((2.93-x)(1.63-x))