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weqwewe [10]
3 years ago
13

For the reaction o(g) + o2(g) → o3(g) δh o = −107.2 kj/mol given that the bond enthalpy in o2(g) is 498.7 kj/mol, calculate the

average bond enthalpy in o3.
Chemistry
1 answer:
Aneli [31]3 years ago
8 0
The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
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<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

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KHSO_4+KOH\rightarrow K_2SO_4+H_2O

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = \frac{1}{1}\times 0.424=0.424moles of potassium sulfate

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Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

0.424mol=\frac{\text{Mass of potassium sulfate}}{174.26g/mol}\\\\\text{Mass of potassium sulfate}=(0.424mol\times 174.26g/mol)=73.88g

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