Question

use standard enthalpies of format to calculate Hrxn for each reaction. 1:2H2S (g) +3O2 (g)     2H2O...

use standard enthalpies of format to calculate Hrxn for each reaction.

1:2H2S (g) +3O2 (g)     2H2O (l) 2SO2 (g)

2: 3NO2 (g)+H2O (l)       2HNO3 (aq) +NO (g)

3: N2O4 (g) + 4H2 (g)      N2 (g) +4H2O (g)

4: CR2O3 (S) +3CO (g)     2CR (S)+3CO2 (g)

Calculate Kc for each reaction and predict whether reactant and product s will be favor at equilibrium

1: CH3OH(g)         CO (g) + 2H2 (g)

2: 2CO2 (g)+2CF2 (g)        4COF2 (g)

3: 6COF2 (g)         3CO2(g)+3CF2 (g)

4: 2CH3OH (g)         2CO (g) +4H2 (g)

Homework Answers

Answer #1

Q1.

HRxn = Hproducts - Hreactants =

1:2H2S (g) +3O2 (g) --> 2H2O (l) + 2SO2 (g)

HRxn = Hproducts - Hreactants

HRxn = 2*H2O + 2*SO2 - (2H2S + 3O2)

HRxn = 2*−285.830 + 2*−296.81 - (2*−20.6 + 3*0)

HRxn = -1124.08 kJ/rxn

2: 3NO2 (g)+H2O (l)    -->    2HNO3 (aq) +NO (g)

HRxn = Hproducts - Hreactants

HRxn = 2*HNO3 + 2*NO - (3NO2 + H2O)

HRxn =2*−174.1 + 2*+90.2 - (3*33.2 + −285.830)

HRxn = 18.43 kJ/rxn

3: N2O4 (g) + 4H2 (g)    -->   N2 (g) +4H2O (g)

HRxn = N2 + 4*H2O - (N2O4 + 4*H2)

elemental N2 = H2 = 0

HRxn = 4*H2O - (N2O4)

HRxn = 4*−241.826 - 9.2

HRxn = -976.504 kJ/rxn

4: Cr2O3 (S) +3CO (g)  -->    2Cr (S)+3CO2 (g)

HRxn = 2*0 + 3*−393.51 -(−1139.7 + 3*−110.53)

HRxn = 290.76 kJ/mol

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