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2 years ago

Which will not appear in the equilibrium constant expression for the reaction below?

Chemistry

1 answer:

n200080 [17]2 years ago

5 0

Answer:

[C] carbon solid

Explanation:

Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.

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Be sure to answer all parts. one of the most important industrial sources of ethanol is the reaction of steam with ethene derive

lions [1.4K]

Answer: 2.17x10⁻³ atm

Explanation:

First, we must write the balanced chemical equation for the process:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

The chemical reactions that occur in a closed container can reach a state of chemical equilibrium that is characterized because the concentrations of the reactants and products remain constant over time. The equilibrium constant of a chemical reaction is the value of its reaction quotient in chemical equilibrium.

The equilibrium constant (K) is expressed as the ratio between the molar concentrations (mol/L) of reactants and products. Its value in a chemical reaction depends on the temperature, so it must always be specified.

We will use the the equilibrium constant Kc of the reaction to calculate partial pressure of ethene. The constant Kc for the above reaction is,

Kc = $\frac{[C_{2} H_{5}OH]}{[H_{2}O][C_{2} H_{4}]}$

According to the law of ideal gases,

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the gas constant (0.082057 atm L / mol K) .

We can use the ideal gas law to determine the molar concentrations ([x] = n / V) from the gas pressures of ethanol and water, assuming that all gases involved behave as ideal gases. In this way,

PV = nRT → P = (n/V) RT → P = [x] RT → [x] = P / RT

So,

$[C_{2} H_{5}OH] = \frac{200 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 4.06 \frac{mol}{L}$

$[H_{2}O] = \frac{400 atm}{0.082057 \frac{atm L}{mol K} x 600 K } = 8.12 \frac{mol}{L}$

So, the molar concentration of ethene (C₂H₄) will be,

$[C_{2} H_{4}] = \frac{[C_{2} H_{5}OH]}{[H_{2}O] x Kc} = \frac{4.06 \frac{mol}{L} }{8.12 \frac{mol}{L}x9.00 x 10^{3} \frac{L}{mol} } = 5.56 x 10^{-5}\frac{mol}{L}$

Then, according to the law of ideal gases,

$P_{C_{2} H_{4}} = [C_{2} H_{4}]RT = 5.56 x 10^{-5} \frac{mol}{L} x 0.082057 \frac{atm L}{mol K} x 600 K = 2.17x10^{-3} atm$

So, when the partial pressure of ethanol is 200 atm and the partial pressure of water is 400 atm, the partial pressure of ethene at 600 K is 2.17x10⁻³ atm.

7 0

1 year ago

If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?

Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

Answer:

0.2 moles O

8 0

2 years ago

Read 2 more answers

A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

IRINA_888 [86]

The mass of a gold bar = 1.424 x 10⁴ gram

<h3>Further Explanation</h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

Known variable

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

Asked

the mass of a gold bar

Answer

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram

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Keywords: density, mass, volume, a gold bar

3 0

2 years ago

Read 2 more answers

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for

joja [24]

Answer:

ΔG° = -118x10³ J/mol

Explanation:

The two half-reactions in the cell are:

Oxidation half-reaction:

Co(s) → Co²⁺(aq) + 2e⁻; E° = -0,28V

Reduction half-reaction:

Cu²⁺(aq)+2e⁻ → Cu(s); E° = 0,34V

The E° of the cell is defined as:

$E_{cell} = E_{red} - E_{ox}$

Replacing:

0,34V - (-0,28V) = 0,62V

It is possible to obtain the keq from E°cell with Nernst equation thus:

nE°cell/0,0592 = log (keq)

Where:

E°cell is standard electrode potential (0.62 V)

n is number of electrons transferred (2 electrons, from the half-reactions)

Replacing:

0,62V×2/0,0592 = log (keq)

20,946 = log keq

keq = 8,83x10²⁰≈ 5,88x10²⁰

ΔG° is defined as:

ΔG° = -RT ln Keq

Where R is gas constant (8,314472 J/molK) and T is temperature (298K):

ΔG° = -8,314472 J/molK×298K ln5,88x10²⁰

ΔG° = -118x10³ J/mol

I hope it helps!

8 0

2 years ago

A protein subunit from an enzyme is part of a research study and needs to be characterized. A total of 0.145 g of this subunit w

Hitman42 [59]

Answer:

The molar mass of the protein is 12982.8 g/mol.

Explanation:

The osmptic pressure is given by:

π=MRT

Where,

M: is molarity of the solution

R: the ideal gas constant (0.0821 L·atm/mol·K)

T: the temperature in kelvins

Hence, we look for molarity:

$0.138 atm=M(0.0821\frac{l*atm}{mol*K} )(28+273K)$

$M=\frac{0.138atm}{(0.0821\frac{l*atm}{mol*K} )(301K)}$ = =5.584×10⁻³mol/l

As we have 2 ml of solution, we can get the moles quantity:

Moles of protein: 5.584×10⁻³ $\frac{mol}{l}\frac{1l}{1000ml}$ ×2ml=1.117×10⁻⁵mol

Finally, the moles quantity is the division between the mass of the protein and the molar mass of the protein, so:

Moles=Mass/Molar mass

Molar mass= Mass/Moles= $\frac{0.145g}{1.117*10^{-5}mol}$ =12982.8 g/mol

8 0

2 years ago