INTRODUCTION - Chapter 6 Thermodynamics class 11 ncert solutions Chemistry - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

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:

ΔU^θof combustion of methane is – X kJ mol^–1. The value of ΔH^θ is

(i) = ΔU^θ

(ii) > ΔU^θ

(iii) < ΔU^θ

(iv) = 0

Answer:

Since ΔH^θ = ΔU^θ + ΔngRT and ΔU^θ = –X kJ mol^–1,

ΔH^θ = (–X) + ΔngRT.

⇒ ΔH^θ < ΔU^θ

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 CH_4 (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