Question
In: Chemistry
Use tabulated DHf data to calculate DH for the reaction:
Use tabulated DHf data to calculate DH for the reaction:
2 NH3(g) + 3 O2(g) + 2 CH4(g) ® 2 HCN(g) + 6 H2O(g)
Solutions
Expert Solution
The reaction is 2 NH3(g) + 3 O2(g) + 2 CH4(g) --> 2 HCN(g) + 6 H2O(g)
ΔHof(NH3(g))= -45.90 kJ/mol, ΔHof(O2(g))= 0, ΔHof(CH4(g))= -74.9 kJ/mol
ΔHof(HCN(g))= +130.5 kJ/mol, ΔHof(H2O(g)) = -241.82kJ/mol
ΔHr= ∑ nΔHof (products)- ∑ nΔHof (reactants)
ΔHr= 2* ΔHof(HCN(g))+6*ΔHof(H2O(g)) - 2*ΔHof(NH3(g)) -3* ΔHof(O2(g))-2* ΔHof(CH4(g))
ΔHr= 2*+130.5 + 6*-241.82 - 2*-45.90 - 2*0 - 2*-74.9 = 261-1450.92 + 91.8 - 0+ 149.8= -948.32 kJ
queen_honey_blossom answered 2 years agoRelated Solutions
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The tabulated data show the concentrations of N2O5 versus time
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Time(s)
[N2O5] (M)
0
1.000
25
0.822
50
0.677
75
0.557
100
0.458
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150
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200
0.210
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First Order: Time vs ln[N2O5]
Second Order: Time vs 1/[N2O5]
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The tabulated data show the concentrations of N2O5 versus time for this reaction: N2O5 (g) -->...
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Time(s)
[N2O5] (M)
0
1.000
25
0.822
50
0.677
75
0.557
100
0.458
125
0.377
150
0.310
175
0.255
200
0.210
1a. Determine the order of the reaction by graphing:
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First Order: Time vs ln[N2O5]
Second Order: Time vs 1/[N2O5]
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1c. Predict the concentration of N2O5 at 250 seconds.
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[X2Y] (M)
0.0
0.100
1.0
0.0856
2.0
0.0748
3.0
0.0664
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0.0598
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Use tabulated half-cell potentials to calculate ΔG∘rxn for each
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Express the energy change in kilojoules to two significant
figures.
Part B O2(g)+2H2O(l)+2Cu(s)→4OH−(aq)+2Cu2+(aq)
Express the energy change in kilojoules to one significant
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Cu2+(aq) + 2 Al(s) --> 3
Cu(s) + 2 Al3+(aq)
II. 6
Cu2+(aq) + 4 Al(s) --> 6
Cu(s) + 4 Al3+(aq)
a. DH°(I) =DH°(II)
b. DH°(I) = 1/2 DH°(II)
c. DH°(I) = 2DH°(II)
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Use the data in the table to calculate the rate of the reaction at
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Experiment
[F2] M
[ClO2] M
Initial rate
(M/s)
1
0.10
0.010
1.2
×
10−3
2
0.10
0.040
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×
10−3
3
0.20
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2.4
×
10−3
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