Chp4_Ktempdep

# Chp4_Ktempdep - Slope = R H ∆-and intercept = R S ∆...

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The temperature dependence of K: S T H G S T H G - = - = 2 2 1 1 assuming S and H are independent of temperature. But: K RT G ln - = So we have: S T H K RT S T H K RT - = - - = - 2 2 1 1 ln ln dividing by –RT: R S RT H K R S RT H K + - = + - = 2 2 1 1 ln ln essentially equation 4.18 on page 114. now subtract: - = - - + - = - 1 2 2 1 1 2 2 1 1 1 ln ln ln ln T T R H K K R S R S RT H RT H K K or - = 1 2 2 1 1 1 ln T T R H K K Note that a plot of ln K vs 1/T yields a straight line with:
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Unformatted text preview: Slope = R H ∆-and intercept = R S ∆ Thus, if rxn is endothermic, slope is negative. As T increases (or 1/T decreases), ln K increases thus favoring the products. If rxn is exothermic, slope is positive. As T increases, ln K decreases thus favoring reactants. Consider this when working Problem 5 of WA Q09....
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