Chp4_Ktempdep - 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

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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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This note was uploaded on 03/29/2009 for the course CH 201 taught by Professor Warren during the Fall '07 term at N.C. State.

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