IMG-5337.JPG Q Search Question 1 10 pts In class, I mentioned that you will need

Question

IMG-5337.JPG Q Search Question 1 10 pts In class, I mentioned that you will need to calculate the Enthalpy and Entropy of four chemical reactions to fit the four reaction types depicted on the Van't Hoff plot, see below. To answer this you may simply email me you work showing the values for the calculated enthalpy and entropy. You do not need to input anything here at this time. Note: if the reactions are blatantly ordering or disordering, you do not need to calculate the entropy using the standard state tables. Inkeq I/T HTML Editor

Explanation / Answer

since deltaG = Gibbs free energy change =-RTlnKeq

Keq= equilibrium constant, deltaG= deltaH-T*deltaS

hencde deltaH-T*deltaS= -RT lnKeq

lnKeq= -deltaH/RT+ deltaS/R. So a plot of lnKeq vs 1/T is straight line whose slope is -deltaH/R and intercept is deltaS/R.

for the given reaction, 2C8H18+ 25O2(l) ---------->16CO2(g)+18H2O(g) is combustion reaction and deltaH is -ve for this reaction and hence deltaH/R is +ve. since 1/T=0 gives the intercept of deltaS/R. for the combustion reaction, there is an increase in no of gaseous moles and an increase in entropy. So entropy is also positive. So the curve A represets this whose slope is -deltaH/R is +ve and intercept deltaS/R is also +ve. ( A is correct).

curve B shows slope of positive deltaH/R but negative deltaS/R. Curve D represents -ve deltaH/R and D also the same,

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