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The temperature of a reaction

  1. Oct 26, 2012 #1
    I'm supposed to find out the temperature of a reaction at equilibrium. I've calculated K=0.187, ΔS0=940J, and I know that ΔG0=1134kJ and ΔH0=1414kJ.

    I'm supposed to use the R*T*ln(K)=ΔG0 formula, but my only problem is that ΔG0 is given for standard conditions,, ie where the temperature equals 298K.

    But I am supposed to use it anyway in the assignment, ie I need to reform it into ΔG0=ΔH0 + T*ΔS0=T*R*ln(K), and calculate T.

    However, ΔS and ΔH for the reaction are given at standard conditions. When the temperature changes they too change, so how is it possible to get a reliable answer?
     
  2. jcsd
  3. Oct 28, 2012 #2
    help?
     
  4. Oct 28, 2012 #3

    AGNuke

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    Gold Member

    Can you post the original question? You might be missing some kind of assumption, like the temperature remains constant for the reaction.
     
  5. Oct 28, 2012 #4
    Here's the reaction. Ca3(PO4)2 (s) + 3 SiO2 (s) + 5 C (s) -> 3 CaSiO3 (s) + 1/2 P4 (g) + 5 CO (g)

    I'm supposed to find the equilibrium temperature, with the data from the OP (all of which are at standard conditions). I am also to assume that none of the substances change their aggregate states (ie none of them turn into gases) going from standard temperature to the equilibrium temperature.
     
  6. Oct 28, 2012 #5

    AGNuke

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    Can you find out K at standard conditions. Using this, relate K and T at standard conditions with K and T asked in the question by Van't Hoff equation.[tex]\mathrm{ln}\frac{K_2}{K_1}=\frac{-\Delta H^{\circ}}{R}\left ( \frac{1}{T_2}-\frac{1}{T_1} \right )[/tex]
     
  7. Oct 29, 2012 #6

    Borek

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    Staff: Mentor

    Do you really need K? Isn't it just a matter of finding T such that [itex]\Delta G = \Delta H - T \Delta S = 0[/itex]?
     
  8. Oct 29, 2012 #7
    AGNuke: I knew K at equil. temperature.. What I was struggling with was assuming that ΔS and ΔH are the same at 298K and 1500K for the reaction... Anyway I feel I understand allot more.

    The assignment wanted that the equilibrum temperature be calculated for K=0.187.. ugh I was probably unclear about this.. sorry i'm sleep deprived as heck.

    EDIT: When I wrote that ΔG0=R*T*ln(K) in the OP, I meant ΔG0=-R*T*ln(K). I forgot the minus-sign infront of R*T*ln(K).
     
    Last edited: Oct 29, 2012
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