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The heat of combustion of methane at 298K is expressed by CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l), ΔH = − 890.2 k J. The magnitude of ΔE of the reaction at this temperature is
  • a)
    Infinity
  • b)
    Less than ΔH
  • c)
    Equal to ΔH
  • d)
    Greater than ΔH
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The heat of combustion of methane at 298K is expressed by CH4 (g) + 2O...
→ CO2 (g) 2H2O (l)

The balanced equation tells us that one molecule of methane (CH4) reacts with two molecules of oxygen (O2) to form one molecule of carbon dioxide (CO2) and two molecules of water (H2O).

The heat of combustion of methane is the amount of heat released when one mole of methane is burned completely in oxygen to form carbon dioxide and water.

The standard heat of combustion of methane is -890.4 kJ/mol. This means that when one mole of methane reacts with oxygen to form carbon dioxide and water, 890.4 kJ of energy is released in the form of heat.

It is important to note that the heat of combustion of methane varies depending on the conditions under which the reaction takes place. For example, if the combustion occurs at a different temperature or pressure, the heat released may be different.
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The heat of combustion of methane at 298K is expressed by CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l), ΔH = − 890.2 k J. The magnitude of ΔE of the reaction at this temperature isa)Infinityb)Less than ΔHc)Equal to ΔHd)Greater than ΔHCorrect answer is option 'B'. Can you explain this answer?
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