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    For the last question, shouldn't it be that the reaction is spontaneous because the change of free energy is negative

    The mark scheme say that's it's not spontaneous. I'm not sure why hmm


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    Lower than 500 K means that you are left of the x-axis intercept i.e. Delta G is positive.
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    (Original post by Pigster)
    Lower than 500 K means that you are left of the x-axis intercept i.e. Delta G is positive.
    What do you mean left of x axis intercept


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    (Original post by Lilly1234567890)
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    For the last question, shouldn't it be that the reaction is spontaneous because the change of free energy is negative

    The mark scheme say that's it's not spontaneous. I'm not sure why hmm


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    ΔG values are on the y axis starting at zero and getting more positive. The line below 500K is always positive in terms of ΔG!
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    As ΔG = ΔH - TΔS, If temperature increases so does the entropy (ΔS) therefore ΔG will inevitably decrease in value regardless of whether ΔH is +/-. As the temperature is increasing, the value of ΔG is decreasing, once the line meets the x axis the reaction becomes feasible as ΔG = 0; therefore reaction unfeasible below 500K as value of ΔG will be > 0.
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    (Original post by Lilly1234567890)
    What do you mean left of x axis intercept
    The line crosses the x-axis at 500 K.

    Temperatures lower than 500 K (i.e. the Q) are to the left of where the line crosses the x-axis.

    At those Ts, the line is above the X-axis. The x-axis starts at 0 and it is convention that values increase on graphs as you move upwards. i.e. Delta G must be positive.
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    Thank you all ! got it


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    (Original post by RME11)
    As ΔG = ΔH - TΔS, If temperature increases so does the entropy (ΔS) therefore ΔG will inevitably decrease in value regardless of whether ΔH is +/-. As the temperature is increasing, the value of ΔG is decreasing, once the line meets the x axis the reaction becomes feasible as ΔG = 0; therefore reaction unfeasible below 500K as value of ΔG will be > 0.
    Careful... the entropic contribution to ∆G becomes more negative with temperature. In this question,*∆S doesn't actually change with temperature.
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    (Original post by KombatWombat)
    Careful... the entropic contribution to ∆G becomes more negative with temperature. In this question,*∆S doesn't actually change with temperature.
    Yes, the value of (T∆S) would have been more appropriate.
 
 
 
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