9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Answers

Answer 1
Answer:

Answer: The value of K_(eq) is 4.66* 10^(-5)

Explanation:

We are given:

Initial moles of ammonia = 0.0120 moles

Initial moles of oxygen gas = 0.0170 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = (0.0120)/(1.00)=0.0120M

Concentration of oxygen gas = (0.0170)/(1.00)=0.0170M

The given chemical equation follows:

                   4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

Initial:          0.0120    0.0170

At eqllm:     0.0120-4x   0.0170-3x       2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 2.20* 10^(-3)M=0.00220M

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.00220\n\n\Rightarrow x=(0.00220)/(2)=0.00110M

Now, equilibrium concentration of ammonia = 0.0120-4x=[0.0120-(4* 0.00110)]=0.00760M

Equilibrium concentration of oxygen gas = 0.0170-3x=[0.0170-(3* 0.00110)]=0.0137M

Equilibrium concentration of water = 6x=[6* 0.00110]=0.00660M

The expression of K_(eq) for the above reaction follows:

K_(eq)=([N_2]^2* [H_2O]^6)/([NH_3]^4* [O_2]^3)

Putting values in above expression, we get:

K_(eq)=((0.00220)^2* (0.00660)^6)/((0.00760)^4* (0.0137)^3)\n\nK_(eq)=4.66* 10^(-5)

Hence, the value of K_(eq) is 4.66* 10^(-5)


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The two reactions above, show routes for conversion of an alkene into an oxirane. If the starting alkene is cis-3-hexene the configurations of the oxirane products, A and B are Product A: _______ Product B: _______ Will either of these two oxirane products rotate the plane of polarization of plane polarized light

Answers

Answer:

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane.

Explanation:

A double bond is converted to an oxirane through oxidation by peracids e.g. mCPBA (meta-chloroperoxybenzoic acid).

Epoxidation can occur at both face of double bond result in formation of two stereoisomers.

Product A and B : (2R,3S)-2,3-diethyloxirane and (2S,3R)-2,3-diethyloxirane

Both A and B contain plane of symmetry. Hence, both the products are achiral. So, they do not rotate the plane of polarization of plane polarized light.

Which expression correctly describes energy using SI units? A. 1 J=1kg•m^2/s^2 B. 1 J= 1kg•m/s^2 C. 1 J= 1kg• m/s D. 1 J= 1kg•m^2/s

Answers

Answer:

A. 1 J=1kg•m^2/s^2

Explanation:

Energy refers to the capacity to do work. According to the International System of units (SI units), energy is measured in Joules.

Energy is represented by the force applied over a distance. Force is measured in Newton (N) and distance in metres (m). Hence, energy is Newton × metre (N.m)

Newton is derived from the SI units of mass (Kilograms), and acceleration (metres per seconds^2) i.e Kg.m/s^2, since Force = mass × acceleration.

Since; Energy = Newton × metres

If Newton = Kg.m/s^2 and metres = m

Energy (J) will therefore be; Kg.m/s^2 × m

1J = Kg.m^2/s^2

How is Hess's law used to measure enthalpy of a desired reaction?A. The enthalpy is obtained from the enthalpy of an intermediate
step.
B. The enthalpy is determined from the enthalpy of similar reactions.
C. The enthalpy from the final equation in a series of reactions is
used
D. Intermediate equations with known enthalpies are added together.

Answers

Hess's law is used to measure the enthalpy of a desired chemical reaction because: D. Intermediate equations with known enthalpies are added together.

What is Hess's Law?

Hess's Law is also known as Hess's law of constant heat summation (enthalpy) and it was named after a Swiss-born Russian chemist called Germain Hess.

Hess's Law states that the energy change (enthalpy) experienced in a desired chemical reaction is equal to the sum of the energy changes (enthalpies) in each chemical reactions that it is made up of or contains.

Read more on Hess's Law here: brainly.com/question/9328637

A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form?Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Answers

Answer:

Mass PbI2 = 18.19 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.19 grams

Answer:

m_(PbI_2)=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_(Pb(NO_3)_2)=(0.14gPb(NO_3)_2)/(1g\ sln)*(1molPb(NO_3)_2)/(331.2gPb(NO_3)_2)  *(1.134g\ sln)/(1mL\ sln) *96.7mL\ sln\n\nn_(Pb(NO_3)_2)=0.04635molPb(NO_3)_2\n\nn_(KI)=(0.12gKI)/(1g\ sln)*(1molKI)/(166.0gKI)  *(1.093g\ sln)/(1mL\ sln) *99.8mL\ sln\n\nn_(KI)=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*(2molKI)/(1molPb(NO_3)_2) =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_(PbI_2)=0.07885molKI*(1molPbI_2)/(2molKI) *(461.01gPbI_2)/(1molPbI_2) \n\nm_(PbI_2)=18.2gPbI_2

Best regards.

Use the internet or your textbook as a reference to compare and contrast the Arrhenius Theory of acids and bases vs. the Brønsted-Lowery Theory. Use the internet or your textbook as a reference to name the following and indicate if they are an acid or a base:
a. HCI
b. KOH
c. HNO
d. Mg(OH),

Answers

Answer and Explanation:

1. Arrhenius Theory which describes the concept protonic. The substance that gives H+ ions when diluted in water is called as an acid (e.g. HCl) and the substance that dissociates OH-ions whenever it is diluted in water is called as the base (e.g. NaOH)

on the other hand

Bronsted Lowery Theory describes the concept of a proton donor-acceptor. The proton-donating species is an acid and the proton-accepting species is known as a base.

2. The Chemical name and nature of acid is shown below:-

Nature Chemical Name

a. HCl Acidic Hydrochloric Acid

b. KOH Basic Potassium hydroxide

c. HNO Acidic Nitric Acid

d. Mg(OH)2 Basic Magnesium hydroxide

wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.572C↽−−⇀DK2=0.226 Calculate the value of the equilibrium constant for the reaction D − ⇀ ↽ − A + 2 B .

Answers

Answer: The value of equilibrium constant for the net reaction is 11.37

Explanation:

The given chemical equations follows:

Equation 1:  A+2B\xrightarrow[]{K_1} 2C

Equation 2:  2C\xrightarrow[]{K_2} D

The net equation follows:

D\xrightarrow[]{K} A+2B

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K=K_1* (1)/(K_2)

We are given:

K_1=2.57

K_2=0.226

Putting values in above equation, we get:

K=2.57* (1)/(0.226)=11.37

Hence, the value of equilibrium constant for the net reaction is 11.37