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    Hi i was just wondering why temperature affects Kc as according to le chateliers principle if temperature is increased then the position of equilibrium will shift to minimise this change, meaning Kc will not change.
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    (Original post by mattvarley)
    Hi i was just wondering why temperature affects Kc as according to le chateliers principle if temperature is increased then the position of equilibrium will shift to minimise this change, meaning Kc will not change.
    The shift in equilibrium position means you will definitely have a different Kc. The shift means that all cincentrations will change.

    The equibrium shifts to minimise the condition change that has been made (e.g. temp or pressure change), it isn't minimising Kc change.
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    (Original post by TutorsChemistry)
    The shift in equilibrium position means you will definitely have a different Kc. The shift means that all cincentrations will change.

    The equibrium shifts to minimise the condition change that has been made (e.g. temp or pressure change), it isn't minimising Kc change.
    why doesn't the equilibrium just shift for a change in temperature restoring the value of Kc
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    (Original post by mattvarley)
    why doesn't the equilibrium just shift for a change in temperature restoring the value of Kc
    Kc changes when the equilibrium shifts.

    May i ask what you understand by Kc?
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    (Original post by mattvarley)
    Hi i was just wondering why temperature affects Kc as according to le chateliers principle if temperature is increased then the position of equilibrium will shift to minimise this change, meaning Kc will not change.
    This is not the easiest concept in A-level to get your head around.

    You probably know that if a reversible reaction is exothermic in one direction, then it is endothermic in the reverse direction. A higher temperature is going to favour the endothermic reaction. An equilibrium (the Kc value if you like) is defined as one where the forward and backward rates are equal. The only way for the equilibrium to have equal rates at a higher temperature is to change the composition of that equilibrium (i.e. new Kc) - so the system does settle back down into equilibrium but with a (slightly) different Kc.
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    (Original post by TutorsChemistry)
    Kc changes when the equilibrium shifts.

    May i ask what you understand by Kc?
    I just don't really understand why it is only temperature that affects the equilibrium constant
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    (Original post by EierVonSatan)
    This is not the easiest concept in A-level to get your head around.

    You probably know that if a reversible reaction is exothermic in one direction, then it is endothermic in the reverse direction. A higher temperature is going to favour the endothermic reaction. An equilibrium (the Kc value if you like) is defined as one where the forward and backward rates are equal. The only way for the equilibrium to have equal rates at a higher temperature is to change the composition of that equilibrium (i.e. new Kc) - so the system does settle back down into equilibrium but with a (slightly) different Kc.
    Oh ok that makes a lot of sense why is it that the equilibrium constant will not change for concentration or pressure
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    Look at an expression of Kc equation for an equilibrium you know. Now think about what happens to the concentration values when you shift the equilibrium right - the product concs on the top become bigger, and the reactant concs on the bottom become smaller. So Kc will increase.
    The opposite will happen if the the eq shjfts left.

    Temperature moves eq one way or the other depending on enthalpy change of the reaction.
    Pressure can also shift eq if there are different moles of gas present on each side of the eq.
    In both cases concentrations will change, so Kc will change.
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    (Original post by mattvarley)
    Oh ok that makes a lot of sense why is it that the equilibrium constant will not change for concentration or pressure
    An equilibrium is unbalanced in terms of energy (enthalpy), so temperature is going to affect one side differently to the other. With pressure and concentration, the differences can be eliminated as you can inter-convert the chemical species on both sides.

    The value of Kc does not alter with changes in concentration. However, the composition of equilibrium does change - i.e. the amounts of each species.This is best illustrated with an example...

    Take a reaction A + B --> C + D where Kc = 4.1. If we have 1 mol of A and B initially, then at equilibrium we would have (ignoring rounding). We can ignore volume here for simplicity as it cancels.
    [A] = 0.33
    [B] = 0.33
    [C] = 0.67
    [D] = 0.67

    Increase the concentration of [A] by adding 1 mol to the mixture, which disturbs the equilibrium. Not currently at equilibrium:
    [A] = 1.33
    [B] = 0.33
    [C] = 0.67
    [D] = 0.67

    As with Le Chatelier's principle, the equilibrium shifts to oppose the change - so it makes more C and D to restore the position (i.e. Kc)
    [A] = 1.15
    [B] = 0.15
    [C] = 0.85
    [D] = 0.85

    The composition of equilibrium is now different (according to Le Chatelier's principle) but Kc remains the same.
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    (Original post by EierVonSatan)
    An equilibrium is unbalanced in terms of energy, so temperature is going to affect one side differently to the other. With pressure and concentration, the differences can be eliminated as you can inter-convert the chemical species on both sides.

    The value of Kc does not alter with changes in concentration. However, the composition of equilibrium does change - i.e. the amounts of each species.This is best illustrated with an example...

    Take a reaction A + B --> C + D where Kc = 4.1. If we have 1 mol of A and B initially, then at equilibrium we would have (ignoring rounding). We can ignore volume here for simplicity as it cancels.
    [A] = 0.33
    [B] = 0.33
    [C] = 0.67
    [D] = 0.67

    Increase the concentration of [A] by adding 1 mol to the mixture, which disturbs the equilibrium. Not currently at equilibrium:
    [A] = 1.33
    [B] = 0.33
    [C] = 0.67
    [D] = 0.67

    As with Le Chatelier's principle, the equilibrium shifts to oppose the change - so it makes more C and D to restore the position (i.e. Kc)
    [A] = 1.15
    [B] = 0.15
    [C] = 0.85
    [D] = 0.85

    The composition of equilibrium is now different (according to Le Chatelier's principle) but Kc remains the same.
    thanks
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    (Original post by TutorsChemistry)
    Look at an expression of Kc equation for an equilibrium you know. Now think about what happens to the concentration values when you shift the equilibrium right - the product concs on the top become bigger, and the reactant concs on the bottom become smaller. So Kc will increase.
    The opposite will happen if the the eq shjfts left.

    Temperature moves eq one way or the other depending on enthalpy change of the reaction.
    Pressure can also shift eq if there are different moles of gas present on each side of the eq.
    In both cases concentrations will change, so Kc will change.
    thanks
 
 
 
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