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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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In which of the following reactions does the heat change represent the heat of formation of water [EAMCET 1991]

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l); Δ H = - 116 k c a l$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); Δ H = - 58 k c a l$

$H^{+} (a q .) + O H^{-} (a q .) \to 2 H_{2} O (l); Δ H = - 13 .7 k c a l$

$C_{2} H_{2} (g) + 2 \frac{1}{2} O_{2} (g) \to 2 C O_{2} (g) + H_{2} O (l)$ ; $Δ H = - 310 k c a l$

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Explanation

Heat of formation is the formation of one mole of the substance from its elements.

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Based on the following thermochemical equations

$H_{2} O (g) + C (s) \to C O (g) + H_{2} (g); Δ H = 131 k J$

$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H = - 282 k J$

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g); Δ H = - 242 k J$

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

The value of X is [CBSE PMT 1992]

–393 kJ

–655 kJ

+393 kJ

+655 kJ

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Explanation

eq. (i) + eq. (ii) + eq. (iii) gives

$X = 131 - 282 - 242 = - 393 k J$ .

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If enthalpies of formation of C₂H₄(g), CO₂(g) and H₂O(l) at 25°C and 1 atm pressure be 52, – 394 and –286 kJ mol^–1 respectively, the enthalpy of combustion of C₂H₄(g) will be [CBSE PMT 1995; AIIMS 1998; Pb. PMT 1999]

+1412 kJ mol^–1

–1412 kJ mol^–1

+141.2 kJ mol^–1

–141.2 kJ mol^–1

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Explanation

$C_{2} H_{4} + 3 O_{2} \to 2 C O_{2} + 2 H_{2} O$

$Δ H_{reaction} = [2 \times Δ H_{f}^{o} (C O_{2}) + 2 \times Δ H_{f}^{o} (H_{2} O)]$ $- [Δ H_{f}^{o} (C_{2} H_{4}) + 3 \times Δ H_{f}^{o} (O_{2})]$

$= [2 (- 394) + 2 (- 286)] - [52 + 0] = - 1412 k J$ .

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Ozone is prepared by passing silent electric discharge through oxygen. In this reaction [AFMC 1998]

Energy is given out

Energy is absorbed

Oxygen is loaded with energy

Oxygen is dissociated into atoms

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Explanation

$3 O_{2} ⇌ 2 O_{3}$ – energy is given out.

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Combustion of glucose takes place according to the equation, $C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$ , $Δ H = - 72 k c a l$ . How much energy will be required for the production of 1.6 g of glucose (Molecular mass of glucose = 180 g) [AFMC 1999]

0.064 kcal

0.64 kcal

6.4 kcal

64 kcal

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Explanation

$Δ H p e r 1 .6 g = \frac{72 \times 1 .6}{180} = 0 .64 k c a l$ .

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