INTRODUCTION - Chapter 6 - Thermodynamics - Ncert Solutions class 11 - Chemistry
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Question 6.1:
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:
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 6.2:
For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0
(ii) Δp = 0
(iii) q = 0
(iv) w = 0
Answer:
A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.
Therefore, alternative (iii) is correct.
Question 6.3:
The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
The enthalpy of all elements in their standard state is zero.
Therefore, alternative (ii) is correct.
Question 6.4:
ΔUθof combustion of methane is – X kJ mol–1. The value of ΔHθ is
(i) = ΔUθ
(ii) > ΔUθ
(iii) < ΔUθ
(iv) = 0
Answer:
Since ΔHθ = ΔUθ + ΔngRT and ΔUθ = –X kJ mol–1,
ΔHθ = (–X) + ΔngRT.
⇒ ΔHθ < ΔUθ
Therefore, alternative (iii) is correct.
Question 6.5:
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:
According to the Question,
Thus, the desired equation is the one that represents the formation of CH4 (g) i.e.,
Enthalpy of formation of CH4(g) = –74.8 kJ mol–1
Hence, alternative (i) is correct.
Question 6.6:
A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Answer:
For a reaction to be spontaneous, ΔG should be negative.
ΔG = ΔH – TΔS
According to the Question, for the given reaction,
ΔS = positive
ΔH = negative (since heat is evolved)
⇒ ΔG = negative
Therefore, the reaction is spontaneous at any temperature.
Hence, alternative (iv) is correct.
Question 6.7:
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Answer:
According to the first law of thermodynamics,
ΔU = q + W (i)
Where,
ΔU = change in internal energy for a process
q = heat
W = work
Given,
q = + 701 J (Since heat is absorbed)
W = –394 J (Since work is done by the system)
Substituting the values in expression (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 6.8:
The reaction of cyanamide, NH2CN(s),with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be –742.7 kJ mol–1at 298 K. Calculate enthalpy change for the reaction at 298 K.
Answer:
Enthalpy change for a reaction (ΔH) is given by the expression,
ΔH = ΔU + ΔngRT
Where,
ΔU = change in internal energy
Δng = change in number of 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 of ΔH:
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
