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3 years ago

In the following reaction, what is the quantity of heat (in kJ) released when 5.87 moles of CH₄ are burned?

1 answer:

5 0

Taking into account the definition of enthalpy of a chemical reaction, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

The enthalpy of a chemical reaction as the heat absorbed or released in a chemical reaction when it occurs at constant pressure. That is, the heat of reaction is the energy that is released or absorbed when chemicals are transformed into a chemical reaction.

The enthalpy is an extensive property, that is, it depends on the amount of matter present.

In this case, the balanced reaction is:

CH₄ (g) + 2 O₂ (g) → CO₂ (g) + 2 H₂O(g)

and the enthalpy reaction ∆H° has a value of -802 kJ/mol.

This equation indicates that when 1 mole of CH₄ reacts with 2 moles of O2, 802 kJ of heat is released.

When 5.87 moles of CH₄ are burned, then you can apply the following rule of three: if 1 mole of CH₄ releases 802 kJ of heat, 5.87 moles of CH₄ releases how much heat?

$heat=\frac{5.87 molesof CH_{4}x802 kJ}{1 mol of CH_{4} }$

heat= 4,707.74 kJ

Finally, the quantity of heat released when 5.87 moles of CH₄ are burned is 4,707.74 kJ.

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Match each substance with the correct designation for the equation HSO3- + CH3NH2 <=> SO32- + CH3NH3+ HSO3- CH3NH2 SO32- C

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Answer:

$HSO_3^-$ : conjugate acid of $SO_3^{2-}$

$CH_3NH_2$ : conjugate base of $CH_3NH_3^+$

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$CH_3NH_3^+$ : conjugate acid of $CH_3NH_2$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

$HSO_3^-+CH_3NH_2\rightleftharpoons SO_3^{2-}+CH_3NH_3^+$

Here in forward reaction $CH_3NH_2$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $CH_3NH_3^+$ which is a conjugate acid.

And $HSO_3^-$ is losing a proton, thus it is considered as an acid and after loosing a proton, it forms $SO_3^{2-}$ which is a conjugate base.

Similarly in the backward reaction, $CH_3NH_3^+$ is loosing a proton, thus it is considered as a acid and after loosing a proton, it forms $CH_3NH_2$ which is a conjugate base.

And $SO_3^{2-}$ is accepting a proton, thus it is considered as a base and after accepting a proton, it forms $HSO_3^{-}$ which is a conjugate acid.

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Explanation:

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