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il63 [147K]
1 year ago
13

Consider the reaction. X ( g ) + Y ( g ) − ⇀ ↽ − Z ( g ) K p = 1.00 at 300 K In which direction will the net reaction proceed fo

r the initial conditions [ X ] = [ Y ] = [ Z ] = 1.0 M? net reaction proceeds to the right net reaction proceeds to the left reaction is at equilibrium In which direction will the net reaction proceed for the initial conditions P X = P Z = 1.0 atm, P Y = 0.50 atm? reaction is at equilibrium net reaction proceeds to the left net reaction proceeds to the right
Chemistry
1 answer:
marta [7]1 year ago
4 0

Answer:

Explanation:

We have in this question the equilibrium

X ( g ) + Y ( g ) ⇆  Z ( g )

With the equilibrium contant Kp = pZ/(pX x pY)

The moment we change the concentration of Y, we are changing effectively the partial pressure of Y since pressure and concentration are directly proportional

pV = nRT ⇒ p = nRT/V and n/V is molarity.

Therefore we can calculate the reaction quotient Q

Qp = pZ/(pX x pY) = 1/ 1  x 0.5 atm = 2

Since Qp is greater than Kp the system proceeds from right to left.

We could also arrive to the same conclusion by applying LeChatelier´s principle which states that any disturbance in the equilibrium, the system will react in such a way to counteract the change to restore the equilibrium. Therefore, by having reduced the pressure of Y the system will react favoring the reactants side increasing some of the y pressure until restoring the equilibrium Kp = 1.

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Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca
lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.23-0.77=-1.0V

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0
1 year ago
At the equivalence point of a KHP/NaOH titration, you have added enough OH- to react with all of the HP- such that the only spec
Nataly_w [17]

Explanation:

The given data is as follows.

  [P^{2-}] = 0.042 M,      K_{a} for HP^{-} = 3.9 \times 10^{-6}

According to the given situation P^{2-} acts as a base.The reaction equation will be as follows.

            P^{2-} + H_{2}O \rightleftharpoons HP^{-} + OH^{-}

Relation between K_{b} and K_{a} are as follows.

                   K_{a} \times K_{b} = K_{w}

                     K_{b} = \frac{1 \times 10^{-14}}{K_{a}}

                                      =  \frac{1 \times 10^{-14}}{3.9 \times 10^{-6}}}

                                      = 2.6 \times 10^{-9}

Also,      K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}

Let us take [OH^{-}] = [HP^{-}] = x

So,                       K_{b} = \frac{[HP^{-}][OH^{-}]}{P^{2-}}

                           2.6 \times 10^{-9} = \frac{x \times x}{0.042}

                      x = 1.04 \times 10^{-5}

[OH^{-}] = [HP^{-}] = 1.04 \times 10^{-5}

                          pOH = - log[OH^{-}]

                                   = - log (1.04 \times 10^{-5})

                                   = 4.99

As it is known that pH + pOH = 14

so,                  pH + 4.99 = 14

                         pH = 9.01

Thus, we can conclude that pH of the solution is 9.01.                  

4 0
1 year ago
How will the following changes affect the mole fraction of chlorine gas, χcl2, in the equilibrium mixture.?
sammy [17]
If the reaction is represented by:
PCl₃ + Cl₂ <-> PCl₅ (exothermic)

the mole fraction of chlorine in the equilibrium mixture will change according to the following:
Decrease the volume: decrease
Increase the temperature: increase
Increase the volume: increase
Decrease the temperature: decrease
4 0
1 year ago
Which of the following represents a pair of isotopes? Select one: a. 146C, 147N b. 11H, 21H c. 136C, 147N d. O2, O3 e. 3216S, 32
skad [1K]

Answer:

B

Explanation:

Isotopy is a phenomenon in which there exists two or more kind of atoms of the same element. These atoms have the same number of protons but different mass numbers. That is, they have the same atomic number but differ only by the number of neutrons in the nucleus of the atom. Now, let us apply this information to the options individually:

A. Is wrong, they are not the same element

B. Is correct. they both belong to the same element hydrogen. Infact, they are named protium and deuterium respectively

C is wrong they are not same element

D is wrong They are not elements at all

E. Is wrong, one is an element, the other is a molecule

3 0
2 years ago
The temperature on a distant, undiscovered planet is expressed in degrees B. For example, water boils at 180 ∘ B and freezes at
marin [14]

Answer:

40.3∘C

Explanation:

At planet B;

Water boils = 180∘C

Water freezes = 50∘C

In this planet the temperature difference = 180 - 50 = 130 compared to earth where the temperature difference is; 100 - 0 = 100

This means;

130 ∘C = 100 ∘C

x ∘C = 31 ∘C

x = 31 * 130 / 100

x = 40.3∘C

5 0
2 years ago
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