Calculate the standard entropy change at 25 degree C for: C_3H_8(g) + 5 O_2(g) r

Question

Calculate the standard entropy change at 25 degree C for: C_3H_8(g) + 5 O_2(g) rightarrow 3 CO_2(g) + 4H_2 O(g) +877.9 J/K +100.5 J/K -72.9 J/K +72.9 J/K +2692 J/K The half-life for the spontaneous decay of iodine-131 is 8.04 days. How much of a 0.50 g sample of this remains after 3.0 days? 0.39 g 0.13 g 0.28 g 0.11 g 0.19 Assume Cu^2+(aq) + Ca(s) rightarrow Cu(s) + Ca^2+(aq) proceeds in the forward direction. We conclude: Ca(s) is oxidized and Cu^2+(aq) is reduced. Cu^2+(aq) is oxidized and Ca(s) is reduced. Ca(s) is oxidized and Ca^2+(aq) is reduced. Cu^2+(aq) is oxidized and Ca^2+(aq) is reduced. Cu(s) is oxidized and Ca(s) is reduced. Calculate the cell potential, at 25 degree C, for 3 Cr^2+(aq) + 2 Al(s) rightarrow 3 Cr(s) + 2 Al^3+(aq). Assume [Cr^2+] = 0.15 M and [Al^3+] = 0.0040 M. +0.73 V +0.44 V -2.64 V +0.77 V -0.75 V For which of the following reactions will the entropy of the system decrease? CaO(s) + CO_2 (g) rightarrow CaCO_3(s) 2 NH_3 (g) rightarrow N_2 (g) + 3 H_2(g) N_2 O_4(g) rightarrow 2 NO_2(g) KOH(s) rightarrow K^+(aq) + OH^-(aq) 2 C(s) + O_2(g) rightarrow 2 CO(g)

Explanation / Answer

7. standard entropy change,

dSo = dSo(products) - dSo(reactants)

       = (3 x 213.68 + 4 x 188.72) - (270.20 + 5 x 205.03)

       = 100.5

Answer : (B) 100.5 J/K

8. rate constant k

k = ln(2)/t1/2 = ln2/8.04 = 0.0862 d-1

amount left [A]

ln[A] = ln[Ao] - kt

         = ln(0.5) - 0.0862 x 3

[A] = 0.39 g

Answer : (A) 0.39 g

9. For forward reaction,

(A) Ca is oxidized, Cu2+ is reduced

10. Ecell = Eo - 0.0592/n logK

               = (-0.91 - (-1.66)) - 0.0592/6 log[(0.004)^2/(0.15)^3]

               = 0.77 V

Answer : (D) 0.77 V

11. Entropy decrease in case of,

(A)

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