All categories

3 years ago

What factors affect the dynamic state of equilibrium in a chemical reaction and how?

Chemistry

1 answer:

yanalaym [24]3 years ago

8 0

Answer:

Only changes in temperature will influence the equilibrium constant $K_c$ . The system will shift in response to certain external shocks. At the new equilibrium $Q$ will still be equal to $K_c$ , but the final concentrations will be different.

The question is asking for sources of the shocks that will influence the value of $Q$ . For most reversible reactions:

External changes in the relative concentration of the products and reactants.

For some reversible reactions that involve gases:

Changes in pressure due to volume changes.

Catalysts do not influence the value of $Q$ . See explanation.

Explanation:

$\displaystyle K_c = {e}^{\Delta G/(R\cdot T)}$ .

Similar to the rate constant, the equilibrium constant $K_c$ depends only on:

$\Delta G$ the standard Gibbs energy change of the reaction, and
$T$ the absolute temperature (in degrees Kelvins.)

The reversible reaction is in a dynamic equilibrium when the rate of the forward reaction is equal to the rate of the backward reaction. Reactants are constantly converted to products; products are constantly converted back to reactants. However, at equilibrium $Q = K_c$ the two processes balance each other. The concentration of each species will stay the same.

Factors that alter the rate of one reaction more than the other will disrupt the equilibrium. These factors shall change the rate of successful collisions and hence the reaction rate.

Changes in concentration influence the number of particles per unit space.
Changes in temperature influence both the rate of collision and the percentage of particles with sufficient energy of reaction.

For reactions that involve gases,

Changing the volume of the container will change the concentration of gases and change the reaction rate.

However, there are cases where the number of gases particles on the reactant side and the product side are equal. Rates of the forward and backward reaction will change by the same extent. In such cases, there will not be a change in the final concentrations. Similarly, catalysts change the two rates by the same extent and will not change the final concentrations. Adding noble gases will also change the pressure. However, concentrations stay the same and the equilibrium position will not change.

You might be interested in

If 21.00 mL of a 0.68 M solution of C6H5NH2 required 6.60 mL of the strong acid to completely neutralize the solution, what was

Andru [333]

Answer:

pH = 2.46

Explanation:

Hello there!

In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

$n_{acid}=n_{base}=n_{salt}$

Whereas the moles of the salt are computed as shown below:

$n_{salt}=0.021L*0.68mol/L=0.01428mol$

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:

$[salt]=0.01428mol/0.0276L=0.517M$

Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

$C_6H_5NH_3^++H_2O\rightleftharpoons C_6H_5NH_2+H_3O^+$

Whose equilibrium expression is:

$Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}$

Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

$2.326x10^{-5}=\frac{x^2}{0.517M}$

Whereas x is:

$x=\sqrt{0.517*2.326x10^{-5}}\\\\x=3.47x10^-3$

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

$pH=-log(3.47x10^{-3})\\\\pH=2.46$

Regards!

6 0

2 years ago

What does RNA do?

MariettaO [177]

Answer:

b, its a polymer that translates genetic information.

7 0

3 years ago

Read 2 more answers

Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W

alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0

3 years ago

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the of strontium-90 is ________ yr.

lesantik [10]

The question is incomplete, here is the complete question.

If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.

a. 28.8 b. 30.9 c. 35.4 d. 32.2

Answer : The half-life of strontium-90 is 28.8 years.

Explanation :

This is a type of radioactive decay and all radioactive decays follow first order kinetics.

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by :

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time taken for decay process = 9.00 year

a = initial amount or moles of the reactant = 1.000 g

a - x = amount or moles left after decay process = 0.805 g

Putting values in above equation, we get:

$k=\frac{2.303}{9.00}\log\frac{1.00}{0.805}$

$k=0.0241\text{ year}^{-1}$

To calculate the half-life, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$0.0241\text{ year}^{-1}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=28.8\text{ years}$

Therefore, the half-life of strontium-90 is 28.8 years.

5 0

3 years ago

What happens at the cathode in an electrolytic cell?<br><br> (Answer is reduction)

melisa1 [442]

Answer:

At the cathode in an electrolytic cell, ions in the surrounding solution are reduced into atoms, which precipitate or plate out on to the solid cathode. The anode is where oxidation takes place, and the cathode is where reduction takes place.

Explanation:

7 0

3 years ago

Read 2 more answers