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alexandr402 [8]
3 years ago
5

How are the following aspects of a reaction affected by the addition of a catalyst? 1) activation energy of the reverse reaction

Chemistry
2 answers:
snow_tiger [21]3 years ago
3 0

These are four questions and four answers.

Answers:

1) activation energy of the reverse reaction

     b. Decreased

2) Rate of the forward reaction

    a. Increased

3) Rate of the reverse reaction

    a. Increased

4) Activation energy of the forward reaction

    b. decreased

Explanation:

<em>Activarion energy</em> is the energy required by the reactants to form the intermediate transition state and become products.

<em>Catalysts</em> are substances that change the path of the chemical reactions, lowering the activation energy, and thus speeding up the rate of the reactions, since the products can reach the new lower activation energy faster.

The equilibrium reactions are the chemical process in which two reactions, the <em>forward and the reverese reactions</em>, occur simultaneously and at the same rate.  The equlibrium reactions may be represented by:

  • A ⇄ B

Where A → B is the direct or forward reaction, and A ← B is the reverse reaction (note the inversed arrow, from right to left).

For the direct reaction A represents the reactants and B represents the products. On the other hand, B represents the reactants and A represents the reactants of the reverse reaction and A. This, is A is the reactant of the forward reaction and the product of the reverse reaction, while B is the reactant of the reverse reaction and the product of the forward reaction.

Since, <em>the addition of a catalyst</em> lowers the activation energy of the process, the new activation energy is lower for both the forward and the reverse reaction, meaning that:

1. <em>The activation energy of the reverse reaction is decreased</em> (option b. of the first question)

2.<em> The rate of the forward reaction is increased</em> (option a. of the second question)

3. <em>The rate of the reverse reaction is increased</em> (option a. of the third question).

4. <em>Activation energy of the forward reaction is decreased</em> (option b. of the fourth question).

In summary, the addition of a catalyst decreases the activation energy for both forward and reverse reactions, and increases the rate of both forward and reverse reactions.

WITCHER [35]3 years ago
3 0

The following aspects of a reaction affected by the addition of a catalyst are

  • 1) activation energy of the reverse reaction: b. Decreased
  • 2) Rate of the forward reaction: a. Increased
  • 3) Rate of the reverse reaction: a. Increased
  • 4) Activation energy of the forward reaction: b. decreased
<h3>Explanation: </h3>

The following aspects of a reaction affected by the addition of a catalyst (it is  a substance that enables a chemical reaction to proceed at a usually faster rate or under different conditions (as at a lower temperature) than otherwise possible) are

1) activation energy of the reverse reaction: b. Decreased

Energy of activation is the energy that must be provided to a chemical or nuclear system with potential reactants to result in: a chemical reaction, nuclear reaction, or other physical phenomena.

2) Rate of the forward reaction: a. Increased

Forward reaction is the reaction that products are produced from reactants and it goes from left to right in a reversible reaction.

3) Rate of the reverse reaction: a. Increased

A reversible reaction is a reaction in which the reactants form products react together to give the reactants back.

4) Activation energy of the forward reaction: b. decreased

The activation energy for the forward reaction is the free energy amount that must be added to go from the energy level of the reactants to the energy level of the transition state

Learn more about catalyst brainly.com/question/1532082

#LearnWithBrainly

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It is given that sodium hydroxide is added to collect the solid nickel(II) hydroxide product

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We will first balance the hydroxide molecule. On the right side there are two OH molecules.

Thus, on the left side we will take 2 sodium hydroxide

Ni2+ + 2NaOH --> Ni (OH)2 + Na+

Now we will balance the sodium ion which are 2 in numbers on the left side and 1 on the right side

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The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

K_a\times K_b=K_w

K_b for sodium formate is K_b=\frac {K_w}{K_a}

Given that:

K_a of formic acid = 1.8\times 10^{-4}

And, K_w=10^{-14}

So,

K_b=\frac {10^{-14}}{1.8\times 10^{-4}}

K_b=5.5556\times 10^{-11}

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HCOONa    ⇒     Na⁺ +    HCOO⁻

Consider the ICE take for the formate  ion as:

                                   HCOO⁻ + H₂O   ⇄   HCOOH + OH⁻

At t=0                              0.35                            -              -

At t =equilibrium           (0.35-x)                          x           x            

The expression for dissociation constant of sodium formate is:

K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}

5.5556\times 10^{-11}=\frac {x^2}{0.35-x}

Solving for x, we get:

x = 0.44×10⁻⁵  M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

<u>pH = 14 - 4.64 = 9.36</u>

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