Chapter 6 Thermodynamics Solutions
Question - 1 : - Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer - 1 : -
A thermodynamic state function is a quantity whose value is independent of a path.
Functions like p, V, T etc. depend only on the state of a system and not on the path.
Hence, alternative (ii) is correct.
Question - 2 : - For the process to occurunder adiabatic conditions, the correct condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0
Answer - 2 : -
A system is said to be under adiabaticconditions if there is no exchange of heat between the system and itssurroundings. Hence, under adiabatic conditions, q = 0.
Therefore,alternative (iii) is correct.
Question - 3 : - The enthalpies of allelements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv)different for each element
Answer - 3 : -
The enthalpy of allelements in their standard state is zero.
Therefore,alternative (ii) is correct.
Question - 4 : - ΔUθof combustion of methaneis – X kJ mol–1. The value of ΔHθ is
(i) = ΔUθ
(ii) > ΔUθ
(iii) < ΔUθ
(iv)= 0
Answer - 4 : -
Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = –X kJmol–1,
ΔHθ = (–X) + ΔngRT.
⇒ ΔHθ < ΔUθ
Therefore,alternative (iii) is correct.
Question - 5 : - The enthalpy of combustion of methane,graphite and dihydrogen at 298 K are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively.Enthalpy of formation of CH4(g) will be
(i) –74.8 kJ mol–1 (ii) –52.27 kJ mol–1
(iii) +74.8 kJ mol–1 (iv) +52.26 kJ mol–1.
Answer - 5 : -
According to the question,
Thus, the desired equation is the one thatrepresents the formation of CH4 (g) i.e.,
Enthalpy of formation of CH4(g) = –74.8 kJ mol–1
Hence,alternative (i) is correct.
Question - 6 : - A reaction, A + B → C + D + q isfound to have a positive entropy change. The reaction will be
(i) possible at hightemperature
(ii) possible only at lowtemperature
(iii) not possible at anytemperature
(iv)possible at any temperature
Answer - 6 : -
For a reaction to be spontaneous, ΔG shouldbe negative.
ΔG = ΔH – TΔS
According to the question,for the given reaction,
ΔS = positive
ΔH = negative (since heat isevolved)
⇒ ΔG = negative
Therefore, the reaction isspontaneous at any temperature.
Hence,alternative (iv) is correct.
Question - 7 : - Ina process, 701 J of heat is absorbed by a system and 394 J of work is done bythe system. What is the change in internal energy for the process?
Answer - 7 : -
According to the first lawof thermodynamics,
ΔU = q + W (i)
Where,
ΔU = change in internal energyfor a process
q = heat
W = work
Given,
q = + 701 J (Since heat is absorbed)
W = –394 J (Since work is done by thesystem)
Substituting the values inexpression (i), we get
ΔU = 701 J + (–394 J)
ΔU = 307 J
Hence,the change in internal energy for the given process is 307 J.
Question - 8 : - The reaction of cyanamide, NH2CN(s),with dioxygen was carriedout in a bomb calorimeter, and ΔU was found to be –742.7kJ mol–1at 298 K. Calculate enthalpy change for thereaction at 298 K.
Answer - 8 : -
Enthalpy change for a reaction (ΔH) is given by theexpression,
ΔH = ΔU + ΔngRT
Where,
ΔU = change in internalenergy
Δng = change in numberof moles
For the given reaction,
Δng = ∑ng (products) – ∑ng (reactants)
= (2 – 1.5) moles
Δng = 0.5 moles
And,
ΔU = –742.7 kJ mol–1
T = 298 K
R = 8.314 × 10–3 kJ mol–1 K–1
Substituting the values in the expression ofΔH:
ΔH = (–742.7 kJ mol–1) + (0.5 mol) (298 K)(8.314 × 10–3 kJ mol–1 K–1)
= –742.7 + 1.2
ΔH = –741.5 kJ mol–1
Question - 9 : - Calculate the number of kJ of heat necessaryto raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heatcapacity of Al is 24 J mol–1 K–1.
Answer - 9 : -
From the expression of heat (q),
q = m. c. ΔT
Where,
c = molar heat capacity
m = mass of substance
ΔT = change in temperature
Substituting the values in the expressionof q:
q = 1066.7 J
q = 1.07 kJ
Question - 10 : - Calculate the enthalpy change on freezing of1.0 mol of water at 10.0°C to ice at –10.0°C. ΔfusH = 6.03 kJ mol–1 at 0°C.
Cp[H2O(l)] = 75.3 J mol–1 K–1
Cp[H2O(s)] = 36.8 J mol–1 K–1
Answer - 10 : -
Total enthalpy changeinvolved in the transformation is the sum of the following changes:
(a) Energy change involvedin the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
(b) Energy change involvedin the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.
(c) Energy change involvedin the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.
= (75.3 J mol–1 K–1) (0 – 10)K + (–6.03 × 103 J mol–1) + (36.8 J mol–1 K–1) (–10 – 0)K
= –753 J mol–1 – 6030 J mol–1 – 368 J mol–1
= –7151 J mol–1
= –7.151 kJ mol–1
Hence,the enthalpy change involved in the transformation is –7.151 kJ mol–1.