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1. The question:

A steel vessel of volume 2.0 dm3 has introduced into it 0.20 mol of SO3, 0.040 mol of SO2 and 0.010 mol of O2.

By calculation of the apparent value of KC show that this mixture is not at equilibrium, and explain in which direction the system will move in order to achieve equilibrium at a temperature of 800K.
The value of KC at this temperature is 1.7 * 10^-6 mol-1 dm3

2SO2(g) + O2 (g) <---> 2SO3

delta H = -196 kJ mol-1

What I have done so far: This is going to look messy

1.7*10^-6 = (SO3)^2/((SO2)^2*(O2))
Do I just substitute the molar values of the components next and see how it compares to the answer in the question?
Also the bit about the temperature doesn't make sense - is it increasing or decreasing?

Thanks for helping, I'm really bad at these type of questions
2. Anyone??
3. (Original post by alde123)
The question:

A steel vessel of volume 2.0 dm3 has introduced into it 0.20 mol of SO3, 0.040 mol of SO2 and 0.010 mol of O2.

By calculation of the apparent value of KC show that this mixture is not at equilibrium, and explain in which direction the system will move in order to achieve equilibrium at a temperature of 800K.
The value of KC at this temperature is 1.7 * 10^-6 mol-1 dm3

2SO2(g) + O2 (g) <---> 2SO3

delta H = -196 kJ mol-1

What I have done so far: This is going to look messy

1.7*10^-6 = (SO3)^2/((SO2)^2*(O2))
Do I just substitute the molar values of the components next and see how it compares to the answer in the question?
Also the bit about the temperature doesn't make sense - is it increasing or decreasing?

Thanks for helping, I'm really bad at these type of questions
All I know is that if a reaction is exothermic like this one, at a higher temperature the position of equilibrium shifts to the right favouring the formation of the product, so 2SO3. Does that help at all? I'm not sure how to explain the other part of the question, sorry.
4. (Original post by Elle_w)
All I know is that if a reaction is exothermic like this one, at a higher temperature the position of equilibrium shifts to the right favouring the formation of the product, so 2SO3. Does that help at all? I'm not sure how to explain the other part of the question, sorry.
Thanks for helping. One of my friends explained it to me at school - turns out I read the question wrong

The temperature is just a stated temperature - it hasn't increased or decreased from anything. The other part of the question was just to prove that it hasn't reached equilibrium yet i.e. just plug the values (moles divided by two) into the Kc expression.

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