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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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For a carnot engine, the source is at 500 K and the sink at 300 K. What is efficiency of this engine [BHU 2004]

0.2

0.4

0.6

0.3

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Explanation

Given that, $T_{1} = 500 K, T_{2} = 300 K$

By using, $η = \frac{T_{1} - T_{2}}{T_{1}} = \frac{500 - 300}{500} = \frac{200}{500} = 0 .4$ .

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From Kirchhoff's equation which factor affects the heat of reaction [MP PMT 1990]

Pressure

Temperature

Volume

Molecularity

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Explanation

Effect of temperature in heat of reaction is given by Kirchoff’s equation.

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The absolute enthalphy of neutralisation of the reaction $M g O (s) + 2 H C l (a q) \to M g C l_{2} (a q) + H_{2} O (l)$ will be [CBSE PMT 2005]

Less than –57.33 kJ mol^–1

–57.33 kJ mol^–1

Greater than –57.33 kJ mol^–1

57.33 kJ mol^–1

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Explanation

Heat of neutralisation will be less than –57.33 kJ/mole because some amount of this energy will be required for the dissociation of weak base (MgO)

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The heat of transition (ΔH_t) of graphite into diamond would be, where

$C (graphite) + O_{2} (g) \to C O_{2} (g); Δ H = x k J$

$C (diamond) + C_{2} (g) \to C O_{2} (g); Δ H = y k J$ [Pb. PET 1985]

$(x + y) k J m o l^{- 1}$

$(x - y) k J m o l^{- 1}$

$(y - x) k J m o l^{- 1}$

None of these

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Explanation

Graphite → diamond $Δ H_{t} = (x - y) k J m o l^{- 1}$ .

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The enthalpy of combustion at 25°C of H₂, cyclohexane (C₆H₁₂) and cyclohexene (C₆H₁₀) are –241, –3920 and –3800 KJ / mole respectively. The heat of hydrogenation of cyclohexene is [BHU 2005]

– 121 KJ / mole

+ 121 KJ / mole

– 242 KJ / mole

+ 242 KJ / mole

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Explanation

(i) $H_{2} + \frac{1}{2} O_{2} \to H_{2} O, Δ H = - 241 k J$

(ii) $C_{6} H_{10} + \frac{17}{2} O_{2} \to 6 C O_{2} + 5 H_{2} O, Δ H = - 3800 k J$

(iii) $C_{6} H_{12} + 9 O_{2} \to 6 C O_{2} + 6 H_{2} O, Δ H = - 3920 k J$

$C_{6} H_{10} + H_{2} \to C_{6} H_{12}$

Eq. (i) + Eq. (ii) – Eq. (iii)

$Δ H = - 241 - 3800 - (- 3920)$

= –4041 + 3920 = – 121 kJ

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The heat change for the reaction $H_{2} + \frac{1}{2} O_{2} \to H_{2} O$ is called

Heat of reaction

Heat of formation

Heat of vaporisation

None of the above

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Explanation

One mole of H₂O is formed from its initial components.

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The heat of neutralisation of a strong acid and a strong alkali is 57.0 kJ mol^–1. The heat released when 0.5 mole of HNO₃ solution is mixed with 0.2 mole of KOH is [KCET 1991; AIIMS 2002;

57.0 kJ

11.4 kJ

28.5 kJ

34.9 kJ

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Explanation

0.2 mole will neutralize 0.2 mole of HNO₃, heat evolved = 57 × 0.2 = 11.4 kJ.

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A solution of 500 ml of 0.2 M KOH and 500 ml of 0.2 M HCl is mixed and stirred; the rise in temperature is T₁. The experiment is repeated using 250 ml each of solution, the temperature raised is T₂. Which of the following is true [EAMCET 1987; MP PET 1994]

T₁ = T₂

T₁ = 2T₂

T₁ = 4T₂

T₂ = 9T₁

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Explanation

Suppose heat evolved in I^st case is Q₁ and that in the II^nd case it is Q₂. Then $Q_{2} = \frac{1}{2} Q_{1 .}$

But $Q_{1} = 1000 T_{1}$ and $Q_{2} = 500 T_{2}$

$∴ 500 T_{2} = \frac{1}{2} \times 1000 T_{1} i . e . T_{2} = T_{1}$ .

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In the reaction for the transition of carbon in the diamond form to carbon in the graphite form, ΔH is –453.5 cal. This points out that [BHU 1981; KCET 1986, 89]

Graphite is chemically different from diamond

Graphite is as stable as diamond

Graphite is more stable than diamond

Diamond is more stable than graphite

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Explanation

$C_{D} \to C_{G}, Δ H = - 453 .5 c a l .$

i.e. energy of C_G is less and thus more stable.

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Which of the following equations correctly represents the standard heat of formation $(Δ H_{f}^{o})$ of methane [IIT JEE (Screening) 1992]

$C (d i a m o n d) + 2 H_{2} (g) = C H_{4} (g)$

$C (g r a p h i t e) + 2 H_{2} (g) = C H_{4} (l)$

$C (g r a p h i t e) + 2 H_{2} (g) = C H_{4} (g)$

$C (g r a p h i t e) + 4 H = C H_{4} (g)$

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Explanation

$C_{(graphite)} + 2 H_{2 (g)} = C H_{4 (g)}$ .

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