(3b) The solubility product of barium iodate, Ba(IO3)2, is 4.01x10-9. Calculate
ID: 591616 • Letter: #
Question
(3b) The solubility product of barium iodate, Ba(IO3)2, is 4.01x10-9. Calculate G. (3c) Consider the following reaction at 0°C: S(s) + 4N0(g) SO2(g) + 2N20(g) A student places 0.025 moles of solid sulfur and 0.100 moles of nitrogen monoxide gas in a closed 1.00 L container at 0oC for several days. Calculate the number of molecules of NO in the 1.00 L container after equilibrium is reached, given the following thermodynamic data: SO2(g)N20(9) 256 297 S(s) S°(J K mol) 31.9 AHo (kJ mor) 0.00 211 90.4 220 81.6Explanation / Answer
3b)
G = -RTlnK
G = -8.314 J/mol.K x 273 K ln(4.01 x 10-9) = 43883.88 J = 43.88 kJ
here temperature is not specified ,but I assume it is 273K , might be given in part a please confirm it
b)
we first have to find out K to find the equilibrium conversion
G = Hrxn-TSrxn
Hrxn = Hfp - Hfr
Hfr = heat of formation of reactants
Hfp = heat of formation of products
Hrxn = 2 x 81.6 + (-297) - {4 x 90.4} = -495.4 kJ = 495400 kJ
Similarly
Srxn = Sfp - Sfr
= 220 x 2 + 256 x 1 -{211 x 4 + 31.9 x 1} = -179.9 J/K
Grxn = Hrxn -TSrxn
Grxn = -495400 J - 273K x (-179.9 J/K) = -446287.3 J
Grxn = -RTlnK
K ~ 1085
reaction is almost complete as k is very very high
NO = zero
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