## butterflydreamer one year ago Equilibrium Constant Calculation (simple one) - please check my working out :)

1. Photon336

Where's the calculation?

2. butterflydreamer

oops sorry, internet cut out. Here it is: In a laboratory test of this equilibrium 6.00 mole of SO2 and 2.50 mole of O2 are added to 2.00L reaction vessel and allowed to come to equilibrium at 700 degrees celsius. If 4.00 mole of SO3 is present at equilibrium, what is the value of the equilibrium constant? $2SO_2 (g) + O_2 (g) <=> 2SO_3 (g)$ So, i drew up an ICE table. |dw:1436876239363:dw| divide by 2 to obtain concentration: |dw:1436876365753:dw| So then the K constant: $K_c = \frac{ [SO_3]^2 }{ [SO_2]^2 \times [O_2]}$ $K_c = \frac{ [2.00]^2 }{ [1.00]^2 \times [0.25] } = 16$ I feel like i'm also meant to refer to the temperature (700 degrees celsius) but i'm not too sure? :/

3. Photon336

The equation doesn't include temperature but it's implied from the value of the equilibrium constant for the reaction, because that will change depending on your temperature.

4. Photon336

I got something around 16 for Kc but I think you need to consider converting to Kp because this is a gaseous reaction since Partial pressures apply.

5. Photon336

There's a formula to convert Kc to Kp and for this you need the temperature. Kp = Kc (RT)^( delta n) Delta n = #moles of gas of product - #moles reactants. R = 0.08 L mol^-1 atm K^-1 T = K

6. butterflydreamer

oh interesting. I haven't learnt about partial pressures yet, but i'll note it down just in case :) thanks.

7. Photon336

Yeah, so for that; the partial pressure In a sealed container the pressure is equal to the sum of the pressures exerted by each individual gas. Which is the mole fraction of that gas multiplied by the total pressure. So if we had a sealed container at some pressure with say two gases A and B one was 2 moles while the other 4; it would be 6 mole of gas total Gas A 2/6 Gas B. 4/6 So pressure total = 2/6*Ptotal+4/6*Ptotal.

8. butterflydreamer

thank you again!! :D