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Anastasy [175]
3 years ago
12

Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O

2(g)→NO2(g), ΔfH∘A=33.2 kJ mol−1 12N2(g)+12O2(g)→NO(g), ΔfH∘B=90.2 kJ mol−1
Chemistry
2 answers:
Alisiya [41]3 years ago
7 0
We apply Hess's Law and obtain:
2ΔHf(B) + ΔH(reaction) = 2ΔHf(A)
ΔH(reaction) = 2(33.2) - 2(90.2)
ΔH(reaction) = -114 kJ / mol
Alika [10]3 years ago
6 0

The enthalpy of the reaction : <u>114 kJ/mol</u>

<h3>Further explanation </h3>

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

Delta H reaction (ΔH) is the amount of heat/heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

ΔH∘rxn = ΔH∘f of the product (s) if ∆Hf ° (reactants) = 0

<em>The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero. </em>

From the problem can be known

∆Hf ° NO₂ = 33.2 kJ / mol

∆Hf ° NO = 90.2 kJ / mol, so that

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)

∆H ° rxn = 2.∆Hf ° NO₂ - 2. ∆Hf ° NO  (O₂ not included)

∆H ° rxn = 2.33.2 - 2.90.2

∆H ° rxn = 66.4 - 180.4

∆H ° rxn = -114 kJ / mol

<h3>Learn more </h3>

Delta H solution

brainly.com/question/10600048

an exothermic reaction

brainly.com/question/1831525

as endothermic or exothermic

brainly.com/question/11419458

an exothermic dissolving process

brainly.com/question/10541336

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Given that a for HBrO is 2. 8×10^−9 at 25°C. What is the value of b for BrO− at 25°C?
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If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

<h3>What is base dissociation constant? </h3>

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 2.8× 10^(-9)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{2.8×10^(-9) }

= 3.5× 10^(-6)

Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).

DISCLAIMER: The above question have mistake. The correct question is given as

Question:

Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?

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7 0
1 year ago
The compound represented by this formula can be classified as an
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It depends what formula you are talking about.

Explanation:

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5 0
3 years ago
Hydrogen gas (H2) reacts with oxygen gas (O2) and produces water. If one mol of hydrogen reacts with one mole of oxygen, what is
Amiraneli [1.4K]
HI 

So, the formula for water is H2O

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3 years ago
Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:
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Answer:

Compound X has a molar mass of 316.25 g*mol^-1 and the following composition:

element & mass %

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sulfur & 60.82%

Write the molecular formula of X.

Explanation:

The given molecule of phosphorus and sulfur has molar mass --- 316.25 g.

Empirical formula calculation:                      

element:              phosphorus                       sulfur

co9mposition:      39.185%                            60.82%

divide with

atomic mass:          39.185/31.0 g/mol           60.82/32.0g/mol

                              =1.26mol                           1.90mol

smallest mole ratio:   1.26mol/1.26mol =1      1.90mol/1.26 mol =1.50

multiply with 2:          2                                         3

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P2S3.

Mass of empirical formula is:

158.0g/mol

Given, molecule has molar mass --- 316.25 g/mol

Hence, the ratio is:

316.25g/mol/158.0 =2

Hence, the molecular formula of the compound is :

2 x (P2S3)

=P_4S_6

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