Explain the relationship between the number of covalent bonds and bond length

The chemistry of nitrogen oxides is very versatile. Given the following reactions and their standard enthalpy changes, (1) NO(g)

AVprozaik [17]

Answer:

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

Explanation:

Given the following reactions and their standard enthalpy changes:

(1) NO(g) + NO₂(g) → N₂O₃(g) ΔH o rxn = −39.8 kJ

(2) NO(g) + NO₂(g) + O₂(g) → N₂O₅(g) ΔH o rxn = −112.5 kJ

(3) 2 NO₂(g) → N₂O₄(g) ΔH o rxn = −57.2 kJ

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

You need to get the heat of reaction from: N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g)

Hess's Law states: "The variation of Enthalpy in a chemical reaction will be the same if it occurs in a single stage or in several stages." That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when verified in a single stage.

This law is the one that will be used in this case. For that, through the intermediate steps, you must reach the final chemical reaction from which you want to obtain the heat of reaction.

Hess's law explains that enthalpy changes are additive. And it should be taken into account:

If the chemical equation is inverted, the symbol of ΔH is also reversed.
If the coefficients are multiplied, multiply ΔH by the same factor.
If the coefficients are divided, divide ΔH by the same divisor.

Taking into account the above, to obtain the chemical equation

N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) you must do the following:

Multiply equation (3) by 2

(3) 2*[2 NO₂(g) → N₂O₄(g) ] ΔH o rxn = −57.2 kJ*2

4 NO₂(g) → 2 N₂O₄(g) ΔH o rxn = −114.4 kJ

Reverse equations (1) and (2)

(1) N₂O₃(g) → NO(g) + NO₂(g) ΔH o rxn = 39.8 kJ

(2) N₂O₅(g) → NO(g) + NO₂(g) + O₂(g) ΔH o rxn = 112.5 kJ

Equations (4) and (5) are maintained as stated.

(4) 2 NO(g) + O₂(g) → 2 NO₂(g) ΔH o rxn = −114.2 kJ

(5) N₂O₅(s) → N₂O₅(g) ΔH o subl = 54.1 kJ

The sum of the adjusted equations should give the problem equation, adjusting by canceling the compounds that appear in the reagents and the products according to the quantity of each of them.

Finally the enthalpies add algebraically:

ΔH= -114.4 kJ + 39.8 kJ + 112.5 kJ -114.2 kJ + 54.1 kJ

ΔH= -22.2 kJ

The heat of reaction for N₂O₃(g) + N₂O₅(s) → 2 N₂O₄(g) is ΔH = -22.2 kJ

8 0

3 years ago

How many moles of oxygen must be placed

umka21 [38]

Answer: 0.245 moles of oxygen must be placed in the container to exert the given pressure at the given temperature. The Ideal Gas Law equation gives the relationship among the pressure, volume, temperature, and moles of gas.

Further Explanation:

The Ideal Gas Equation is:

$PV = nRT$

where:

P - pressure (in atm)

V - volume (in L)

n - amount of gas (in moles)

R - universal gas constant $0.08206 \frac{L-atm}{mol-K}$

T - temperature (in K)

In the problem, we are given the values:

P = 2.00 atm (3 significant figures)

V = 3.00 L (3 significant figures)

n = ?

T = 25.0 degrees Celsius (3 significant figures)

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:

$Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15$

Therefore, for this problem,

$Temperature\ in \ K = 25.0 +273.15\\Temperature\ in \ K = 298.15$

Solving for n using the Ideal Gas Equation:

$n \ = \frac{PV}{RT}\\n \ = \frac{(2.00 \ atm) \ (3.00 \ L)}{(0.08206 \ \frac{L-atm}{mol-K})( 298.15 \ K)} \\n \ = 0.245 \ mol$

The least number of significant figures is 3, therefore, the final answer must have only 3 significant figures.

Learn More

1. Learn more about Boyle's Law brainly.com/question/1437490

2. Learn more about Charles' Law brainly.com/question/1421697

3. Learn more about Gay-Lussac's Law brainly.com/question/6534668

Keywords: Ideal Gas Law, Volume, Pressure

4 0

3 years ago

A watershed is the area of land where all of the water drains off and eventually combines at a central point. As water runs off

Ivanshal [37]

Answer:

The answer is C (or 3 in this case)

Explanation:

I took the test and got it correct and the explanation is that the watershed will flow into more water which spreads that pollution and as it flows could pick up more pollution.

3 0

3 years ago

Read 2 more answers

If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after t

Naya [18.7K]

Answer is: the mass of the excess reactant (ethane) leftover is 90.135 grams.
Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g).
m(C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.

6 0

4 years ago

Read 2 more answers

Concentration required to begin precipitate pbcl2 for pbcl2 ksp=1. 17×10−5

Ne4ueva [31]

The concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.

<h3>What is molarity?</h3>

Molarity is the measure of the concentration of any solute in per unit volume of the solution.

The reaction is $\rm Pb^2^+(aq) + 2Cl^-(aq) \rightarrow PbCl_2(s).$

The molarity of lead is 0.025 M

The ksp is given 17×10⁻⁵

Now, calculating the concentration

$[Pb^2^+] = 0.025 M.\\Ksp = 1.17 \times 10^-^5\\Ksp = [Pb^2^+] \times [Cl^-]^2\\[Cl^-] = \dfrac{\sqrt{ Ksp}}{[Pb^2^+]} \\\\[Cl^-] = \dfrac{\sqrt{ 0.0000117}}{0.025} \\[Cl^-] = 2.16 \times 10^-^2M.$

Thus, the concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.

Learn more about molarity

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5 0

2 years ago