Question

At 87°C 22.0 miles of X (g) , 33.0 moles of Y2 (g) and 4.0 moles...

At 87°C 22.0 miles of X (g) , 33.0 moles of Y2 (g) and 4.0 moles of X3Y5 (g) are mixed in a 80.0 liter container. At equilibrium 10.0 moles of the product gas are present. Calculate Kc and Kp     6X (g) + 5 Y2 (g) --> 2 X3Y5 (g)

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Homework Answers

Answer #1

Step 1. Create an ICE table:

6 X (g) 5 Y2 (g) ------> 2 X3Y5 (g)
initial 22 mol 33 mol 4 mol
change -18 mol = -6 x 3mol -15 mol = -5 x 3mol + 6 mol = 2 x 3 mol
equilibrium = 4 mol = 18 mol 10 mol

Step 2. Find molarities in equilibrium:

[X] = 4.0 moles / 80.0 L = 0.05 M

[Y2] = 18.0 moles / 80.0 L = 0.225 M

[X3Y5] = 10.0 moles / 80 L = 0.125 M

Step 3. Set up the equation and solve for Kc:

Step 4. Set up the equation and solve for Kp:

T = 87 °C = 360.15 K

n = ( 2 ) - ( 6+5 ) = -9

R = 0.0821 L atm / mol K

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