H2O) = 0512.get, mol K\'\', h = 6.626 x 10\'a Js, c = 3.0 x 102 m/s, 760 mm hen

ID: 1030716 • Letter: H

Question

H2O) = 0512.get, mol K'', h = 6.626 x 10'a Js, c = 3.0 x 102 m/s, 760 mm hen ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The decomposition produces vapor, and chromium(l) oxide. The temperature is constant at 25 nitrogen gas, water °C [-1S points total] Substance Cr Os(s) H20(g) N2(g) S" (J/mol K) 0.08121 0.1187 0.1915 0.1137 -1147 -242 -22.5 a Write a balanced chemical reaction for the decomposition of ammonium dichromate b. c. 2 Determine G"(in kJ/mol). Determine K for this reaction. 2cg) 2 sCs) 2092.4-(298) (0.00062181 let) G"-1-2092 4. Balance the following redox reaction in BASIC solution: [-10 points]

Explanation / Answer

(a)

(NH4)2Cr2O7(s) -----> N2(g) + 4H2O(g) + Cr2O3(s)

(b)

dGo = dHo – TdSo

T = 25 oC = 298 K

Substance dHfo(KJ/mol) So(KJ/mol K) dGo = dHo – TdSo

Cr2O3(s)

-1147

0.08121

-1171.20058

H2O(g)

-242

0.1187

-277.3726

N2(g)

0.1915

-57.067

(NH4)2Cr2O7(s)

-22.5

0.1137

-56.3826

So, dGo = dGo(products) - dGo(reactants)
              = [ (-57.067) + 4 (-277.3726) + (-1171.20058) ] – (-56.3826)
              = -2281.37538 kJ/mol
              = -2281 kJ/mol

(c)

We know that

dGo = - R T ln k

So, substituting the values, we get;

-2281 kJ mol-1 = - (8.314 J K-1 mol-1) x (298 K) x ln K
-2281000 J mol-1 = - 2477.572 J mol-1 x ln K
ln K = (-2281000 J mol-1) / (- 2477.572 J mol-1)
ln K = 920.66
ln K = 921
k = exp(921) =