Equilibrium - 2020 Q3aiii

Chemistry Level 2 2.6 Chemical reactivity
1

Hello

I am struggling with question 3aiii from the 2020 paper (Chemical Reactivity).
Here’s a screenshot:

image
image1761×1005 122 KB

My response was:
50<400
The lower value of Kc shows that there are fewer products and more reactants present. This means the reverse reaction has taken place.

I didn’t understand the answer in the marking schedule. What is ‘K?’ And how do we calculate it?

If there are any other gaps you see in my understanding, please feel free to point them out. Always eager to learn more :slight_smile:
Thanks

2
  • Kc is the equilibrium constant of the reaction which is basically a ratio of the concentration of reactants and the concentration of products in an equilibrium at a given time.
  • This can be calculated using Kc = [products] divided by [reactants], where [products] is the concentration of the products of the reaction (which in this case are CF4 and CO2) and [reactants] is the concentration of reactants in the equation (in this example the reactant is COF2.)
  • The number of moles for a reactant or product are written as a power in the equilibrium expression, so because there are 2 moles of COF2 (which we can tell because it is written as (2COF2) we write it as [COF2]^2 in the equilibrium expression.
  • We can then use this equilibrium expression from Qa (i) to answer question (ii) - all we need to do is substitute the values we have been given for the concentrations of the reactants and products into the equation, giving us the value of Kc at this particular time.
  • For question (iii), you need to know that when the value of Kc is greater than 1, the concentration of products is greater than the concentration of reactants so the equilibrium has favoured the products, and when the value of Kc is less than 1, the concentration of reactants is greater than the concentration of products so the equilibrium has favoured the reactants.
  • In this case because the Kc value is 50 which is greater than 1, the concentration of products is still greater than the concentration of reactants so the products are still favoured (even though it has decreased from the previous temperature where Kc was 400 which is probably why you thought the reverse reaction has been favoured).
  • I’ve attached an image which helps with understanding forming an equation and calculating Kc.
    image
3

Thanks for that. It makes sense

In Q3iii, the marking scheme referred to ‘K.’ Is this the same thing as ‘Kc?’ Or is it something different?

4

K is a generic term which represents any equilibrium constant, while Kc is the value of the equilibrium constant in terms of concentration of products and reactants - in this context I think they are pretty much the same thing though

5

Good response Emilie. But I would note a couple of things.
One is that the idea that K=1 is the cutoff for whether products or reactants are favoured is only a rough guideline as the maths changes when exponents need to be taken into account. So a more realistic guidline is that values of K> 100 will probably favour products and values of K< 0.01 will probably favour the reactants
Because if this the answer provided to the exam question is not as good as it can be. Instead it should have stated that because at the new tenperature the value of K has DECREASED from 200 to 50, this means the equilibrium position has shifted to favour the reactants.