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Define standard enthalpy of formation. - Chemistry

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प्रश्न

Define standard enthalpy of formation.

व्याख्या

उत्तर

  1. The standard enthalpy of formation of a compound is the enthalpy change that accompanies a reaction in which one mole of pure compound in its standard state is formed from its elements in their standard states.
  2. Consider the reaction, \[\ce{H2_{(g)} + 1/2 O2_{(g)} -> H2O_{(l)}}\], ΔrH° = −286 kJ
    When one mole of liquid water is generated from H2 and O2 gases in their standard states, the enthalpy changes correspond to the standard enthalpy of water creation.
    ΔfH° of water is −286 kJ mol−1
  3. Similarly, the formation of one mole of CH4 from carbon and hydrogen in their standard states is depicted as:

\[\ce{C_{(graphite)} + 2H2_{(g)} -> CH4_{(g)}}\]

ΔrH° = −74.8 kJ or 

ΔfH (CH4)

= −74.8 kJ mol−1

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Thermochemistry
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पाठ 4: Chemical Thermodynamics - Short answer questions (Type- II)

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एससीईआरटी महाराष्ट्र Chemistry [English] 12 Standard HSC
पाठ 4 Chemical Thermodynamics
Short answer questions (Type- II) | Q 4

संबंधित प्रश्‍न

Answer the following question.

State Hess’s law of constant heat summation. Illustrate with an example. State its applications.


Calculate the total heat required

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b) heat it to 100 °C and then

c) vapourise it at that temperature.

[Given: ΔfusH° (ice) = 6.01 kJ mol-1 at 0 °C, ΔvapH° (H2O) = 40.7 kJ mol-1 at 100 °C, Specific heat of water is 4.18 J g-1 K-1]


The standard enthalpy of formation of water is - 286 kJ mol-1. Calculate the enthalpy change for the formation of 0.018 kg of water.


Calculate the standard enthalpy of combustion of CH4(g) if ΔfH°(CH4) = – 74.8 kJ mol–1, ΔfH°(CO2) = – 393.5 kJ mol–1 and ΔfH°(H2O) = – 285.8 kJ mol–1.


Calculate the standard enthalpy of formation of liquid methanol from the following data:

  1. \[\ce{CH3OH_{(l)} + \frac{3}{2} O_{2(g)} -> CO_{2(g)} + 2H2O_{(l)}}\]     ∆H° = – 726 kJ mol–1
  2. \[\ce{C_{(Graphite)} + O_{2(g)} -> CO_{2(g)}}\]          ∆cH° = – 393 kJ mol–1
  3. \[\ce{H_{2(g)} + \frac{1}{2} O_{2(g)} -> H2O_{(l)}}\]          ∆fH° = – 286 kJ mol–1 

Calculate the standard enthalpy of the reaction.

\[\ce{2Fe_{(s)} + \frac{3}{2} O_{2(g)} -> Fe2O_{3(s)}}\]

Given:

1. \[\ce{2Al_{(s)} + Fe2O_{3(s)} -> 2Fe_{(s)} + Al_2O_{3(s)}}\], rH° = –847.6 kJ
2. \[\ce{2Al_{(s)} + \frac{3}{2} O_{2(g)} -> Al2O_{3(s)}}\], rH° = –1670 kJ

Does the following reaction represent a thermochemical equation?

\[\ce{CH_{4(g)} + 2O_{2(g)} -> CO_{2(g)} + 2H2O_{(g)}}\], ∆fH° = –900 kJ mol–1


When 2 moles of C2H6(g) are completely burnt, 3129 kJ of heat is liberated. If ∆Hf for CO2(g) and H2O(l) are −395 and −286 kJ per mole respectively, the heat combustion of C2H6(g) is ____________.


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Enthalpy of formation of two compounds x and y are −84 kJ and −156 kJ respectively. Which of the following statements is CORRECT?


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\[\ce{C6H12O6_{(s)} + 6O2_{(g)} -> 6CO2_{(g)} + 6H2O_{(g)}}\]; ΔH = −72 kcal mol−1

The energy needed for the production of 1.8 g of glucose by photosynthesis will be ___________.


Standard enthalpy of formation of water is - 286 kJ mol-1. When 1800 mg of water is formed from its constituent elements in their standard states the amount of energy liberated is ______.


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Heat of formation of water is - 272 kJ mol-1. What quantity of water is converted to H2 and O2 by 750 kJ of heat?


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\[\ce{N2H_{4(g)} + H_{2(g)} -> 2NH_{3(g)}}\]

If ΔH0(N – H) = 389 kJ mol–1, ΔH0(H – H) = 435 kJ mol–1, ΔH0(N – N) = 159 kJ mol–1.


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ΔcH0[C2H5OH(1)] = - 1409 kJ mol-1


Standard enthalpy of combustion of a substance is given. Then Write thermochemical equation.

ΔcH0[CH3CHO(l)] = - 1166 kJ mol-1


The enthalpy of combustion of S (rhombic) is − 297 kJ mo1-1. Calculate the amount of sulphur required to produce 29. 74 kJ of heat.


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