In: Chemistry
3. a) Using only the following information:
• ∆H°f for NO (g) is +90.4 kJ/mol
• ∆H° = –56.6 kJ/mol for the reaction: NO (g) +
1/2 O2 (g) à NO2 (g)
Determine ∆H°f for NO2 (g).
b) Using only your answer to (a) and the following information:
•∆H° = –283.0 kJ/mol for the reaction: CO (g) + 1/2 O2 (g) à CO2 (g) Determine ∆H° for the reaction: 4 CO (g) + 2 NO2 (g) à 4 CO2 (g) + N2 (g)
c) A 10.0-L vessel contains 5.0 atm of CO and 3.0 atm of NO2 at
25°C. How much heat (in Joules) will be liberated if this is
allowed to react to completion according to the reaction in part
(b)?
d) In a separate experiment using a very large reaction vessel with
a movable piston, 6.00 moles of CO2 (g) reacts completely with 3.00
moles of nitrogen gas according to the following equation at 25°C
and with
a constant external pressure of 2.00 atm:
4 CO2 (g) + N2 (g) à 4 CO (g) + 2 NO2
(g)
This reaction proceeds to completion. Calculate ∆U, q,
and w for this reaction under these conditions.
3 a) Determine ΔH°f for NO2 (g).
ΔHrxn = ΔH°f(NO2(g)) – ΔH°f(NO(g)) – ½ ΔH°f(O2(g))
–56.6 kJ/mol = ΔH°f(NO2(g)) – 90.4 kJ/mol – ½ (0)
Solve for ΔH°f(NO2(g)) = 33.8 kJ/mol
b) We can write the overall reaction as the sum
of two reactions using Hess’s Law:
4 CO (g) + 2 O2 (g) → 4 CO2 (g) ΔH° = 4(–283.0)
2 NO2 (g) → 2 O2 (g) + N2
(g)
ΔH° = –2(ΔH°f(NO2)) = –67.6
4 CO (g) + 2 NO2 (g) → 4 CO2 (g) + N2 (g) ΔH° = –1199.6 kJ/mol
c) From the balanced equation 4 CO (g) + 2 NO2
(g) → 4 CO2 (g) + N2 (g) , we need
twice as much CO as NO2. But we only have 5 atm of CO for the 3 atm
of NO2, and thus we can
see that CO is the limiting reagent and will all react.
Initially, we have
n =PV/RT
=(5.0 atm)(10.0 L)/(0.0821L atm/K mol)(298 K)
= 2.04 mol CO
Find the amount of heat liberated:
2.04 mol CO ×(1 mol reaction/4 mol CO)×1199.6 kJ/mol reaction×1000
J/1 kJ
= 6.1 × 105 J
d) A quick comparison will show us that CO2 (g)
is the limiting reagent and nitrogen is present in excess.
Also notice that this reaction is the reverse of the reaction from
parts (b) and (c), so we need to switch
the sign on ΔH°.
q = nΔH°
=6.00 mol CO2 /4 * 1199.6 kJ/mol = 1799.4 kJ
w = –P ΔV
We need to calculate Vinitial and Vfinal :
Vinitial = (ninitial)RT / P = (9.00 moles)(0.0821)(298
K) / (2.00 atm) = 110.0 L
To find Vfinal we need to know nfinal, so
let’s do a RICE table:
R:
4 CO2 (g) + N2 (g) --------> 4 CO (g)
+ 2 NO2 (g)
I:
6.00
3.00
0
0
C:
–4(1.50)
–1.50
+4(1.50)
+2(1.50)
E:
0
1.50
6.00
3.00
So nfinal= 0 + 1.50 + 6.00 + 3.00 = 10.50
moles
Vfinal = (nfinal)RT / P = (10.50 moles)(0.0821)(298 K) / (2.00 atm)
= 128.4 L
ΔV = Vfinal – Vinitial = 128.4 L – 110.0 L = 18.4 L
so: w = –P ΔV = –(2.00 atm)(18.4 L) = – 36.8 L•atm
–36.8 L•atm × (101.3 J)/(1 L•atm) = –3727 J = – 3.73 kJ
ΔU = q + w = 1799.4 kJ + – 3.73 kJ = 1795.7 kJ
ΔU = 1795.7 kJ
q = 1799.4 kJ
w = – 3.73 kJ