61 6 x 83145 29815 x ln1061 kj rg29815k

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Unformatted text preview: 19.4 kJ f) 20.4 kJ g) 27.8 kJ h) 31.2 kJ Show your calculations! In this problem, we consider the above chemical reaction under two different conditions 1) at 298.15K and 10.6 bars and 2) at 298.15K and 1.0 bar. We make a cycle based on this observation between: Reactants (298.15K and 10.6 bars) à༎ Products (298.15K and 10.6 bars) (1)â༎ á༎(3) Reactants (298.15K and 1 bar) à༎ Products (298.15K and 1 bar) (2) ΔRG(298.15K and 10.6 bars) = ΔG1(reactants @298.15K from 10.6 bar to 1 bar) + ΔRG2 (reaction @ 298.15K and 1 bar) + ΔG3(products @298.15K from 1 bar to 10.6 bar) ΔG1(reactants @298.15K from 10.6 bar to 1 bar) = n(H2(g)) RT ln (1/10.6) + 5 n(I2(s)) Vm(I2(s)) (1 – 10.6) bar x 10 Pa/bar Note that the change in the free energy for iodine is written as V ΔP because iodine is considered an incompressible solid. Molar volumes of solids and liquids are much smaller than gases, so the term V ΔP is negligible in comparison to the nRT ln(Pf/Pi) term. ΔG1(reactants @298.15K from 10.6 bar to 1 bar) = 3 x 8.3145 x 298.15 x ln (1/10.6) kJ ΔRG2 (reaction @ 298.15K and 1 bar) = 3 mol x (2 x 1.7 kJ/mol) = 10.2 kJ ΔG3(products @298.15K from 1 bar to 10.6 bar) = n(HI(g)) RT ln (10.6/1) = 6 x 8.3145 * 298.15 x ln(10.6/1) kJ ΔRG(298.15K and 10.6 bars) = 27.8 kJ Question 6: (5 points) Consider the following chemical reaction at 1 bar: l2 (s) + H2 (g) à༎ 2HI (g) Using the data in the data section and assuming that the reaction enthalpy change and the rea...
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