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Ivanshal [37]
3 years ago
5

Suppose that the gas-phase reactions A→B and B→A are both elementary processes with rate constants of 4.7×10−3s−1 and 5.8×10−1s−

1, respectively.
What is the value of the equilibrium constant for the equilibrium A(g)⇌B(g)?
Chemistry
1 answer:
uranmaximum [27]3 years ago
8 0

Answer:

K = 8.1 x 10⁻³

Explanation:

We are told here that these gas phase reactions are both elementary processes, thus the reactions forward and reverse are both first order:

A→B        Rate(forward) = k(forward) x [A]

and for

B→A       Rate(reverse) = k(reverse) x [B]

At equilibrium we know the rates of the forward and reverse reaction are equal, so

k(forward) x [A]  = k(reverse) x [B]    for A(g)⇌B(g)

⇒  k(forward) / k(reverse)  =  [B] / [A] =  K

4.7 x 10⁻³  s⁻1 / 5.8 x 10⁻¹ s⁻¹ = 8.1 x 10⁻³ = K

Notice how this answer is logical : the rate of the reverse reaction is greater than the forward reaction ( a factor of approximately 120 times) , and will be expecting a number for the equilibrium constant, K, smaller than one where the reactant concentration, [A], will prevail.

It is worth to mention that this is only valid for reactions which are single, elementary processes and not true for other equilibria.

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Predict the products of La(s) + O2(aq) ->
mina [271]

Answer:

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

Explanation:

<em>Hello </em><em>there!</em>

When you are given such a problem for completing the chemical equations, what you have to understand is that metals are found in groups I, II and III. While Oxygen is a group VI element.

From the above question I have considered that my La(s), solid is either Sodium (Na) - group I, Magnesium - group II and Aluminum - group III.

In a reaction, there is exchange of electrons given by their oxidation numbers (I, II and III - for our metals above)

The chemical equations are thus;

i. Na (s) + O2 (g) => NaO (s)

ii. Mg (s) + O2 (g) => Mg2O (s)

iii. Al (s) + O2 (g) => Al2O3 (s)

Relate this to the problem and it will be;

i. La (s) + O2 (g) => LaO2 (s)

ii. La (s) + O2 (g) => La2O (s)

III. La (s) + O2 (g) => La2O3 (s)

<em>I hope this </em><em>helps </em><em>you</em><em> </em><em>to </em><em>understand</em><em> </em><em>better</em><em>.</em><em> </em><em>Enjoy </em><em>your</em><em> </em><em>studies</em>

3 0
2 years ago
Day 2 code word
bagirrra123 [75]

Answer:

chloroplast

Explanation:

4 0
3 years ago
What part of cellular respiration is affected by arsenic
LenaWriter [7]

Answer:

It effects all of the cellular respiration process

Explanation:

It inhibits the Glycolysis. It replaces the phosphate groups that is needed for making Pyruvate and ATP.

5 0
3 years ago
For each of the following pairs of complexes, identify which one you would predict to have the larger Δo value, and explain why.
mash [69]

Answer:

a) [Fe(H2O)6]3+

b) [Fe(CN)6]3−

c) [Ru(CN)6]3-

Explanation:

. [Mn(H2O)6]2+ or [Fe(H2O)6]3+

The both complexes are d5 complexes with the same ligand , water. Water is a weak ligand and note that Mn^2+ often have a crystal field stabilization energy of zero hence

[Fe(H2O)6]3+ will possess a greater ∆o value.

The splitting of d orbitals according to the crystal field theory depends on the;

i)geometry of the complex

ii) nature of the metal ion,

iii)charge on the metal ion,

iv) ligands that surround the metal ion.

When the geometry and the ligands are held constant, the order of crystal field splitting is as follows;

Pt4+ > Ir3+ > Rh3+ > Co3+ > Cr3+ > Fe3+ > Fe2+ > Co2+ > Ni2+ > Mn2+

[Fe(H2O)6]3+ or [Fe(CN)6]3−

[Fe(CN)6]3− will have a greater ∆o because the cyanide ion is a strong field ligand compared to water. A strong field ligand causes a greater splitting of the octahedral crystal field compared to a weak field ligand.

. [Fe(CN)6]3− or [Ru(CN)6]3-

[Ru(CN)6]3- will exhibit a greater crystal field splitting. Crystal field splitting increases with the second and third row transition elements when compared to the crystal field splitting of the first row transition elements. Note that, there is an increase of approximately 30%–50% in Δo on going from a first-row transition metal to a second-row metal and another 30%–50% increase on going from a second-row to a third-row metal when they have the same geometry and oxidation state.

4 0
3 years ago
The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure
Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate \Delta H_{vap} of the reaction, we use clausius claypron equation, which is:

\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

P_1 = vapor pressure at temperature T_1

P_2 = vapor pressure at temperature T_2

\Delta H_{vap} = Enthalpy of vaporization  

R = Gas constant = 8.314 J/mol K

1) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =363 K

T_2 = final temperature =373 K

P_2=1 atm, P_1=?

Putting values in above equation, we get:

\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]

P_1=0.69671 atm \approx 0.6970 atm

The vapor pressure at temperature 363 K is 0.6970 atm

2) \Delta H_{vap}=40.68 kJ/mol=40680 J/mol

T_1 = initial temperature =373 K

T_2 = final temperature =383 K

P_1=1 atm, P_2?

Putting values in above equation, we get:

\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]

P_2=1.4084 atm \approx 1.410 atm

The vapor pressure at 383 K is 1.410 atm

8 0
3 years ago
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