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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.

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At 25°C, K = 0.090 for the following reaction. H2O(g) + Cl2O(g) equilibrium reaction arrow 2 HOCl(g) Calculate the concentration

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Answer:

[HOCl] = 0.00909 mol/liter

[H₂O] = 0.03901 mol/liter

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Explanation:

1. Chemical reaction:

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

2. Initial concentrations:

i) 1.3 g H₂O

Number of moles = 1.3g / (18.015g/mol) = 0.07216 mol

Molarity, M = 0.07216 mol / 1.5 liter = 0.0481 mol/liter

ii) 2.2 g Cl₂O

Number of moles = 2.2 g/ (67.45 g/mol) = 0.0326 mol

Molarity = 0.0326mol / 1.5 liter = 0.0217 mol/liter

3. ICE (Initial, Change, Equilibrium) table

$H_2O(g)+Cl_2O(g)\rightleftharpoons 2 HOCl(g)$

I 0.0481 0.0326 0

C -x -x +x

E 0.0481-x 0.0326-x x

4. Equilibrium expression

$K_c=\dfrac{[HOCl]^2}{[H_2O].[Cl_2O]}$

$0.09=\dfrac{x^2}{(0.0481-x)(0.0326-x)}$

5. Solve:

$x^2=0.09(x-0.0481)(x-0.0326)\\\\0.91x^2+0.007263x-0.000141125=0$

Use the quadatic formula:

$x=\dfrac{-0.007263\pm \sqrt{(0.007263)^2-4(0.91)(-0.000141125)}}{2(0.91)}$

The positive result is x = 0.00909

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[HOCl] = 0.00909 mol/liter

[H₂O] = 0.0481 - 0.00909 = 0.03901 mol/liter

[Cl₂O] = 0.0326 - 0.00909 = 0.02351 mol/liter

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