All categories

2 years ago

Use the bond energies to answer the question.

Chemistry

1 answer:

Schach [20]2 years ago

4 0

Answer:

Explanation:

Hello!

This is a topic I just learned this year so I might not help the best as of explaining. But I will try my best.

To solve this problem, we must first figure the balanced equation.

H₂+O₂=H₂O₂

Now, we have to find the bond energy of the reactants and products. This is how.

Let's find the reactant first.

For the compound H₂, the Lewis dot structure says that it only consists of one singular bond. So the bond energy of H₂ would be 432.

For the compound O₂, the Lewis dot structure that is the best out of the resonance structures form a double bond. Let's use the bond energy 495 for this since there are double bonds and we need the corresponding double bond energy to make the bond energy accurate.

Let's add this amount together. 927

Now, we have to find the bond energy of the products.

H₂O₂ sounds complicated but it's fairly easy. When you write the Lewis dot structure for this specific compound, you will get something along the lines of this:

H-O-O-H (disregarding lone pairs)

In this compound, there are two bonds of H-O and one bond that is a O-O. This might take you a long time to see this, but just think of what atom is touching the atom next door. This might help distinguish the bonds there are between the compound.

Let's take the bond energy of H-O (467) and multiply this by two since we have two. Add that number to the bond energy of O-O (146). We should have the numbers 1080.

Since the problem is asking for the difference in total energy between the reactants and products, let's subtract the total bond energy of the products-reactants.

1080-927=153

Therefore, your final answer of the total bond energy is 153.

You might be interested in

Write a balanced equation for the single-replacement oxidation-reduction reaction described, using the smallest possible integer

jeyben [28]

Answer:

$\rm Cl_2 + 2\; KBr \to Br_2 + 2\; KCl$ .

One chlorine molecule reacts with two formula units of (aqueous) potassium bromide to produce one bromine molecule and two formula units of (aqueous) potassium chloride.

Explanation:

<h3>Formula for each of the species</h3>

Start by finding the formula for each of the compound.

Both chlorine $\rm Cl$ and bromine $\rm Br$ are group 17 elements (halogens.) Each
On the other hand, potassium $\rm K$ is a group 1 element (alkaline metal.) Each

Therefore, the ratio between $\rm K$ atoms and $\rm Br$ atoms in potassium bromide is supposed to be one-to-one. That corresponds to the empirical formula $\rm KBr$ . Similarly, the ratio between

The formula for chlorine gas is $\rm Cl_2$ , while the formula for bromine gas is $\rm Br_2$ .

<h3>Balanced equation for the reaction</h3>

Write down the equation using these chemical formulas.

$\rm ?\; Cl_2 + ?\; KBr \to ?\;Br_2 + ?\; KCl$ .

Start by assuming that the coefficient of compound with the largest number of elements is one. In this particular equation, both $\rm KBr$ and $\rm KCl$ features two elements each.

Assume that the coefficient of $\rm KCl$ is one. Hence:

$\rm ?\; Cl_2 + 1 \; KBr \to ?\;Br_2 + ?\; KCl$ .

Note that $\rm KBr$ is the only source of $\rm K$ and $\rm Br$ atoms among the reactants of this reaction.

There would thus be one $\rm K$ atom and one $\rm Br$ atom on the reactant side of the equation.

Because atoms are conserved in a chemical equation, there should be the same number of $\rm K$ and $\rm Br$ atoms on the product side of the equation.

In this reaction, $\rm Br_2$ is the only product with $\rm Br$ atoms.

One $\rm Br$ atom would correspond to $0.5$ units of $\rm Br_2$ .

Similarly, in this reaction, $\rm KCl$ is the only product with $\rm K$ atoms.

One $\rm K$ atom would correspond to one formula unit of $\rm KCl$ .

Hence:

$\displaystyle \rm ?\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl$ .

Similarly, there should be exactly one $\rm Cl$ atom on either side of this equation. The coefficient of $\rm Cl_2$ should thus be $0.5$ . Hence:

$\displaystyle \rm \frac{1}{2}\; Cl_2 + 1 \; KBr \to \frac{1}{2}\;Br_2 + 1\; KCl$ .

That does not meet the requirements, because two of these coefficients are not integers. Multiply all these coefficients by two (the least common multiple- LCM- of these two denominators) to obtain:

$\displaystyle \rm 1\; Cl_2 + 2 \; KBr \to 1\;Br_2 + 2\; KCl$ .