Question

Consider the reaction:
4CO(*g*)+2NO2(*g*)?4CO2(*g*)+N2(*g*).

Using the following information, determine ?H? for the reaction
at 25?C.

NO(g) ?H?f = +91.3 kJ/mol

CO2(g) ?H?f = -393.5 kJ/mol

2NO(g)+O2(g)?2NO2(g) ?H? = -116.2 kJ/mol

2CO(g)+O2(g)?2CO2(g) ?H? = -566.0 kJ/mol

Express your answer using one decimal place.

Answer #1

Look at the last equation: 2CO(g)+O2(g) -----> 2CO2 (g) delta H= -566

delta H = (heat of formation of products) - (heat of formation
of reactants)

delta H = (2 x heat of formation of CO2) - (2 x heat of formation
of CO + heat of formation of O2)

-566 kJ = (2 x -393.5 kJ) - (2 x heat of formation of CO) + 0 since
the heat of formation of an element is 0

from this, heat of formation of CO = -111 kJ

Look at the third equation: 2NO(g)+O2(g)-----> 2NO2(g) delta H= -116.2

delta H = (2 x heat of formation of NO2) - (2 x heat of
formation of NO + heat of formation of O2)

-116.2 kJ = (2 x heat of formation of NO2) - (2 x 91.3) + 0

from this, heat of formation of NO2 = 33.2 kJ

Now look at the given equation: 4CO(g)+2NO2(g) ---> 4CO2(g)+N2(g)

delta H = (4 x heat of formation of CO2) - (4 x heat of
formation of CO) - (2 x heat of formation of NO2)

= (4 x -393.5 kJ) - (4 x -111 kJ) - (2 x 33.2 kJ) = -1196 kJ

Hess’s Law b) Calculate the ∆H for the reaction:
Ti(s) + 2Cl2 (g) → TiCl4 (l)
Using the following chemical equations and their respective
enthalpy changes:
Ti(s) + 2Cl2 (g) → TiCl4 (g) ∆H = -763 kJ TiCl4 (l) → TiCl4 (g)
∆H = 41 kJ
b) Calculate the ∆H for the reaction:
2CO(g) + O2 (g) → 2CO2 (g)
Using the following chemical equations and their respective
enthalpy changes:
2C(s) + O2 (g) → 2CO(g) ∆H = -221.0 kJ...

For the reaction 2NO(g) + O2(g) → 2NO2(g),
∆H° = -113.1 kJ/mol and ∆S° = -145.3 J/K·mol. Under which
temperature conditions would the reaction be spontaneous?

Consider the following reaction and corresponding value of Kc
2NO(g)+O2(g)⇌2NO2(g)Kc=5.4×1013 at 25∘C What is the value of Kp at
this temperature? Express your answer using two significant
figures.

Using the information below determine the change in enthalpy for
the following reaction:
2NO(g) + 5 H2(g) → 2NH3(g) +
2H2O(l)
H2(g) + ½ O2(g) →
H2O(l)
∆H°= -285.8kJ
N2(g) + O2(g) →
2NO(g)
∆H°= +180.5 kJ
2NH3(g) → N2(g) +
3H2(g)
∆H°= + 92.22 kJ
please explain step by step! thanks!

Calculate the standard enthalpy of formation (∆H°f) at 298.15 K
for CO(g)
Known: Reaction ∆H°R298.15 K,
cal/mol for C(gr) + O2(g) →
CO2(g) = -94,052 cal/mol
and 2CO(g) + O2(g) → 2CO2(g) =
-135,272

Calculate the enthalpy change for the following reaction, in kJ
mol-1
N2O4(g) + Cl2(g)
→ 2NOCl(g) + O2(g)
given the following
data: ∆H
(kJ mol-1)
2NOCl(g) → 2NO(g) +
Cl2(g) +75.56
2NO(g) + O2(g) →
2NO2(g) -113.05
2NO2(g) →
N2O4(g
) -58.03

Calculate ΔG∘ (in kJ/mol) for the following reaction at 1 atm
and 25 °C:
C2H6 (g) + O2 (g) →
CO2 (g) + H2O (l) (unbalanced)
ΔHf C2H6 (g) = -84.7 kJ/mol; S
C2H6 (g) = 229.5 J/K⋅mol;
ΔHf ∘ CO2 (g) = -393.5 kJ/mol; S
CO2 (g) = 213.6 J/K⋅mol;
ΔHf H2O (l) = -285.8 kJ/mol; SH2O
(l) = 69.9 J/K⋅mol;
SO2 (g) = 205.0 J/K⋅mol

Consider the reaction
N2(g) +
2O2(g)----->2NO2(g)
Using the standard thermodynamic data in the tables linked above,
calculate Grxn for this reaction at 298.15K if the
pressure of each gas is 29.25 mm Hg.
Consider the reaction
4HCl(g) +
O2(g)------>2H2O(g)
+ 2Cl2(g)
Using the standard thermodynamic data in the tables linked above,
calculate G for this reaction at 298.15K if the pressure of each
gas is 12.40 mm Hg.

Use the information in the table below to determine
DGofor the reaction
2NH3(g) + 2O2(g) --> N2O(g)
+ 3H2O(l)
DGo (kJ)
N2(g) + 3H2(g) -->
2NH3(g)
-33.0
4NH3(g) + 5O2(g) --> 4NO(g) +
6H2O(l)
-1010.0
N2(g) + O2(g) --> 2NO(g)
174.9
N2(g) + 2O2(g) -->
2NO2(g)
102.6
2N2(g) + O2(g) -->
2N2O(g)
204.2

Using the Information below determine the change in enthalpy for
the following reaction:
2NO (g) + 5H2
(g)!2NH3
(g) + 2H2O
(l)
H2 (g) + 1⁄2O2 (g)!H2O (l) ΔH° = -285.8 kJ
N2 (g) + O2 (g)!2NO (l) ΔH° = +180.5 kJ
2NH3 (g)!N2 (g) + 3H2 (g)ΔH° = +92.22 kJ
a)-197.52 kJ
b)-241.7 kJ
c)-483.3 kJ
d)-659.88 kJ
e)-844.3 kJ
please show me which equation is first second and thrid and
reason why? i may be taking the wrong...

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