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Sphinxa [80]
3 years ago
11

Use bond energies to calculate ΔHrxn Δ H r x n for the reaction. 2H2(g)+O2(g)→2H2O(g) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( g )

Chemistry
1 answer:
Olenka [21]3 years ago
3 0

Answer:

\large \boxed{\text{-486 kJ}}

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

                      2H₂   +   O₂ ⟶ 2H-O-H

Bonds:          2H-H    1O=O       4H-O

D/kJ·mol⁻¹:     436      498          464

\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.

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What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?
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Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g 
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Let us use the above mentioned data (in g) and values will be ​​converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

C:  \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol

H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

C:  \dfrac{6.975}{6.975} = 1

H:  \dfrac{16.3}{6.975} \approx 2.3

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark

I hope this helps. =)
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