Engineering Chemistry S Datta

THERMODYNAMICS 93
Example 2. The amount of work of a system is 8.36 × 10 9 ergs, after absorbing 990 cals.
of heat. Calculate the internal energy change of the system.
Sol. We have,
∆E = q – w
Here q = 990 cals =
∴
990 × 4.2
10 3
w = 8.36 × 10 9 ergs =
kJ = 4.13
8.36 × 10
10 × 10
9
7 3
∆E = 4.158 – 0.836 = 3.332 kJ.
kJ = 0.836 kJ
Example 3. Calculate the work done by the following reaction at 27°C.
Zn + H 2
SO 4
→ ZnSO 4
+ H 2
↑
1 mol.
Sol. Here, w = P(V 2
– V 1
) = PV 2
[Q V 1
= 0]
= nRT = RT [Q n = 1]
= 8.314 × (273 + 27) =
= 2.4942 kJ.
8.314 × 300
1000
Example 4. Calculate the heat of combustion at constant pressure of carbon. (Given q v
at 200°C is 97000 cals).
Sol.
C(s) + O 2
(g) = CO 2
(g)
Here q v
= – 97000 cal.
mols of gaseous reactant, n 1
= 1
mols of gaseous product, n 2
= 1
∴ ∆n = n 2
– n 1
= 0.
We have,
q p
= q v
+ ∆nRT.
= – 97000 + 0 × RT
= – 97000 cals.
So, q p
= q v
(As n 2
= n 1
)
Example 5. Calculate the value of ∆E for the reaction at 500 K, Mg(s) + 2HCl(g) →
MgCl 2
(s) + H 2
(g) (Given ∆H = – 109 Kcal).
Sol. We know, ∆H = ∆E + ∆nRT.
Here,
∆n = 1 – 2 = – 1, T = 500 K
R = 2 cal deg –1 mol –1 = 2 × 10 –3 Kcal deg –1 mol –1
∴ ∆E = ∆H – ∆nRT = – 109 – (– 1) × 2 × 10 –3 × 500
= – 108 Kcal.
Example 6. Heats of combustion of ethylene, hydrogen, and ethane are 337.2, 68.3, 372.8
Kcal mol –1 at a definite temperature. Calculate the heat of reaction of C 2
H 4
(g) + H 2
(g) ⎯⎯→
C 2
H 4
(g) at that temperature.
kJ.