lecture 08 2010

# lecture 08 2010 - 40 Factors affecting equilibrium We...

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40 Factors affecting equilibrium We learned last time that Thus, K eq = e - Δ G ° /RT = e -( Δ H ° - T Δ S ° )/RT = e - Δ H ° /RT × e + Δ S ° /R We can define the enthalpic and entropic factors of the equilibrium constant as Enthalpic factor = and the entropic factor is given as Let’s evaluate both of these terms for the different dissolution processes. For NH 4 NO 3 : Remember that this process is endothermic – the beaker gets cold - Δ H ° /RT = -28000/[(8.31)(298)] = -11.23 e - Δ H ° /RT = e -11.23 = 1.3 × 10 -5 . The enthalpic factor in the equilibrium constant is small. Endothermic reactions, all other factors being equal, are expected to have small K eq . However, the entropic term is important: Δ S ° /R = 108/8.31 = +13.00 e Δ S ° /R = 4.4 × 10 5 . So, K eq = (1.3 × 10 -5 )(4.4 × 10 5 ) = 5.7 So, the favorable entropic term make the dissolution of NH 4 NO 3 occur. The favorable entropy arises from the fact that the ordered crystalline structure of the solid is replaced by the more disordered solution. That is enough to overcome the unfavorable heat effect

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41 For NaOH: Remember that this process is exothermic – the beaker gets hot. - Δ H ° /RT = +43000/[(8.31)(298)] = +17.36 e - Δ H ° /RT = e +17.36 = 3.1 × 10 7 . The enthalpic factor in the equilibrium constant is very large. Exothermic reactions, all other factors being equal, are expected to have large K eq . Now, let’s consider the entropic term.
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## This note was uploaded on 02/13/2010 for the course CHM 29275 taught by Professor Farrar during the Spring '10 term at Rochester.

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lecture 08 2010 - 40 Factors affecting equilibrium We...

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