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nlexa [21]
2 years ago
9

The heat of combustion for sugar is 5639 kJ/mole, but that for TNT is 3406 kJ/mole. Why, if the heat of combustion for TNT is sm

aller than for sugar, is TNT an explosive?
Chemistry
1 answer:
n200080 [17]2 years ago
7 0

Heat of combustion used to explain the relative stability of the compounds.

The heat of combustion definition is the variation in energy released in the form of heat through the burning of an amount of any substance, all reactants being in the standard state (temperature of 25 C and pressure of 1 atm).

When the amount is quantified for 1 mole of said substance, one can define this quantity as the molar heat of combustion.

The molar heat of combustion values are defined in kilojoules per mole (KJ/mol). The kilojoule is a unit of energy and corresponds to 1000 or Joules (J). Mole is a unit of measurement used to express the amount of matter.

Now sugar is a non electrolyte so comparatively more stable hence more energy required to decompose the sugar hence its heat of combustion value is -5639kJ/mol high.

Where as TNT used as an explosive so it is susceptible for reaction and combustion, hence low amount of heat of combustionnis required for TNT that is -3406kJ/mol whuch is comparatively lower.

To know more about heat of combustion here

brainly.com/question/11103047

#SPJ4

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Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,
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The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: K_{c} = 2.10^{-2}

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CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = \frac{20}{125} = 0.16M

<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

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