Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. For such system

(i) qp will be negative

(ii) ∆_rH will be negative

(iii) qp will be positive

(iv) ∆_rH will be positive

Thermodynamics mainly deals with

(i) Interrelation of various forms of energy and their transformation from one form to another.

(ii) Energy changes in the processes which depend only on initial and final states of the microscopic systems containing a few molecules.

(iii) How and at what rate these energy transformations are carried out.

(iv) The system in equilibrium state or moving from one equilibrium state to another equilibrium state.

Which of the following is not correct?

(i) ∆G is zero for a reversible reaction

(ii) ∆G is positive for a spontaneous reaction

(iii) ∆G is negative for a spontaneous reaction

(iv) ∆G is positive for a non-spontaneous reaction

Enthalpy of sublimation of a substance is equal to

(i) Enthalpy of fusion + enthalpy of vapourisation

(ii) Enthalpy of fusion

(iii) Enthalpy of vapourisation

(iv) Twice the enthalpy of vapourisation

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

(i) Is always negative

(ii) Is always positive

(iii) May be positive or negative

(iv) Is never negative

Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct?

(a) C (g) + 4H (g) → CH₄ (g); ∆r H = x kJ/ mol

(b) C (graphite,s) + 2H₂ (g) → CH₄ (g); ∆rH = y kJ /mol

(i) x = y

(ii) x = 2y

(iii) x > y

(iv) x < y

On the basis of thermochemical equations (a), (b) and (c), find out which of the algebric relationships given in options (i) to (iv) is correct.

(a) C (graphite) + O₂ (g) → CO₂ (g) ; ∆_rH = x kJ/ mol

(b) C (graphite) + $\frac{1}{2}$ O₂ (g) → CO (g) ; ∆_rH = y kJ/ mol

(c) CO (g) $\frac{1}{2}$ +  O2 (g) → CO₂ (g) ; ∆_rH = z kJ/ mol

(i) z = x + y

(ii) x = y – z

(iii) x = y + z

(iv) y = 2z – x

The entropy change can be calculated by using the expression ∆S = . When water freezes in a glass beaker, choose the correct statement amongst the following:

(i) ∆S (system) decreases, but ∆S (surroundings) remains the same.

(ii) ∆S (system) increases, but ∆S (surroundings) decreases.

(iii) ∆S (system) decreases, but ∆S (surroundings) increases.

(iv) ∆S (system) decreases, and ∆S (surroundings) also decreases.

The pressure volume work for an ideal gas can be calculated by using the expression w = - ${\int }_{{\mathit{V}}_{\mathit{i}}}^{{\mathbf{V}}_{\mathbf{f}}}{\mathit{P}}_{\mathit{e}\mathit{x}}$ dv. The work can also be calculated from the pV−plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf. Choose the correct option.

(i) w (reversible) = w (irreversible)

(ii) w (reversible) < w (irreversible)

(iii) w (reversible) > w (irreversible)

(iv) w (reversible) = w (irreversible) + p_ex.?V

In an adiabatic process, no transfer of heat takes place between system and surroundings. Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following.

(i) q = 0, ∆T ≠ 0, w = 0

(ii) q ≠ 0, ∆T = 0, w = 0

(iii) q = 0, ∆T = 0, w = 0

(iv) q = 0, ∆T < 0, w 0 V_f