lec3_2011

# lec3_2011 - Applications of the vant Hoff equation Provided...

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Applications of the van’t Hoff equation Provided the reaction enthalpy,Δ r H θ , can be assumed to be independent of temperature, eqn. 7.23b (or 9.26b in 7 th edition) illustrates that a plot of –ln K against 1/T should yield a straight line of slope Δ r H θ /R. Example: The data below show the equilibrium constant measured at different temperatures. Calculate the standard reaction enthalpy for the system. T/K 350 400 450 500 K 3.94x10 -4 1.41x10 -2 1.86x10 -1 1.48 Solution: 1/T 2.86x10 -3 2.50x10 -3 2.22x10 -3 2.00x10 -3 -lnK 7.83 4.26 1.68 -0.39

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Continued
Example: The equilibrium constant of the reaction 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g) is 4.0x10 24 at 300K, 2.5x10 10 at 500K, and 2.0x10 4 at 700K. Estimate the reaction enthalpy at 500K. Solution: discussion: 1. Do we need a balanced reaction equation here? 2. What can be learned about the reaction based on the information provided? 3. Will the enthalpy become different at 300K or 700K?

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Calculate the value of K at different temperatures The equilibrium constant at a temperature T 2 can be obtained in terms of the know equilibrium constant K 1 at T 1 . Since the standard reaction enthalpy is also a function of temperature, when integrating the equation 9.26b from T 1 to T 2 , we should assume that Δ r H ө is constant within that interval. so ln( K 2 ) – ln( K 1 ) = (7.24) Equation 7.24 provides a non-calorimetric method of determining standard reaction enthalpy. (Must keep in mind that the reaction enthalpy is actually temperature-dependent ) ( ) ln( T d R H K d K K T T r 1 2 1 2 1 1 1 - = θ ) 1 1 2 1 ( T T R H r - -
, The Haber reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) At 298 K, the equilibrium constant K = 6.1x10 5 . The standard enthalpy of formation for NH 3 equals -46.1 kJ mol -1 . What is the equilibrium constant at 500K? Answer: First, calculate the standard reaction enthalpy, Δ r H ө , Δ r H ө = 2*Δ f H ө (NH 3 ) - 3* Δ f H ө (H 2 ) - Δ f H ө (N 2 ) = 2*(-46.1) – 3*0 - 1*0 = - 92.2 kJ mol -1 then ln( K 2 ) – ln(6.1*10 5

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lec3_2011 - Applications of the vant Hoff equation Provided...

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