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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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If 1.00 kcal of heat is added to 1.2 L of oxygen in a cylinder at constant pressure of 1.000 atm, the volume increases to 1.5 L, Hence $∆ E$ for this process is:

0.993 kcal

1.0073 kcal

0.0993 kcal

1.00073 kcal

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Explanation

$∆ H = ∆ E + P ∆ V$

1.00 = $∆ E + \frac{1 (1.5 - 1.2) L a t m}{0.082 L a t m} \times 2 \times 10^{- 3} k c a l ∆ E = 0.993 k c a l$

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Heats of combustion of $C H_{4}, C_{2} H_{4}, C_{2} H_{6}$ are -890, -1411 and -1560 kJ/mole respectively. Which has the lowest fuel value in kJ/gm ?

$C H_{4}$

$C_{2} H_{4}$

$C_{2} H_{6}$

All same

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Explanation

The heat of combustion of per gm of $C H_{4}$

= $- \frac{890}{16} = - 55.625 k J$

The heat of combustion of per gm of $C_{2} H_{4}$

= $\frac{1411}{28} = - 50.393 k J$

The heat of combustion of per gm of $C_{2} H_{6}$

= $- \frac{1560}{30} = - 52 k J$

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The standard heat of combustion of a hydrocarbon compound is an/a-

Extensive property

Colligative property

Intensive property

Constitutive property

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Explanation

The standard heat of combustion is defined as the heat evolved on burning one mole of the compound in presence of O₂ to give CO₂ and water. Since the amount of the substance is specified, it becomes an intensive property.

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The Kirchhoff's equation gives the effect of....on heat of reaction.

Pressure

Temperature

Volume

Molecularity

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Explanation

Kirchhoff's equation is:

$∆ H_{2} - ∆ H_{1} = ∆ C_{P} (T_{2} - T_{1})$

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The temperature of 5 ml of a strong acid increases by $5^{\circ}$ when 5 ml of a strong base is added to it. If 10 ml of each is mixed, temperature should increase by-

$5^{\circ}$

$10^{\circ}$

$15^{\circ}$

cannot be known

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Explanation

When 5 ml of strong acid reacting with 5 ml of strong base net, heat evolved = $∆ H$

$∴$ temperature increase = $\frac{∆ H}{m s}$

when 10 ml of each are reaction, the heat evolved = $2 ∆ H;$

but mass of the solution = 2m

$∴$ temperature increase = $\frac{2 ∆ H}{2 m s} = \frac{∆ H}{m s}$

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The heat of neutralisation of HCl by NaOH is -55.9 kJ/mole. If the heat of neutralisation of HCN by NaOH is -12.1 kJ/mole, then energy of dissociation of HCN is-

-43.8 kJ

43.8 kJ

68 kJ

-68 kJ

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Explanation

The heat of neutralisation = -55.9 kJ

$∴$ The heat of neutralisation of HCN by NaOH = the heat of neutralisation + the energy of dissociation of HCN

= -12.1 kJ

or the energy of dissociation of HCN

= +55.9-12.1 = 43.8 kJ

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For a reaction at equilibrium-

$∆ G = ∆ G^{\circ} = 0$

$∆ G = 0 b u t n o t ∆ G^{\circ}$

$∆ G^{\circ} = 0 b u t n o t ∆ G$

$∆ G = ∆ G^{\circ} \neq 0$

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Explanation

For a reaction at equilibrium $∆ G = 0 a n d n o t ∆ G^{\circ} .$

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Using only the following data:

(I) $F e_{2} O_{3} (s) + 3 C O (g) ⇌ 2 F e (s) + 3 C O_{2} (g); ∆ H^{\circ} = - 26.8 k J$

(II) $F e (s) + C O (g) ⇌ F e O (s) + C O (g); ∆ H^{\circ} = + 16.5 k J$

the $∆ H^{\circ}$ value, in kilojoules, for the reaction

$F e_{2} O_{3} (s) + C O (g) \to 2 F e O (s) + C O_{2} (g)$ is calculated to be:

-43.3

-10.3

+6.2

+10.3

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Explanation

from equation (I) + (2 $\times$ II) ; $∆ H^{\circ} = 6.2 k J$

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82 litres of carbon dioxide are produced at a pressure of 1 atm by the action of acid on a metal carbonate. The work done by the gas (in calories) in pushing back the atmosphere is (R = 0.082 litre-atm deg^-1mol^-1)

1000

820

1640

2000

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Explanation

The work done by a gas is given by

W= $P ∆ V$

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Among the following, the reaction for which $∆ H = ∆ E$ is-

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$

$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$

$C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)$

$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$

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Explanation

When $∆ n = 0, ∆ H = ∆ E$ because $∆ H = ∆ E + R T ∆ n .$

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