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Crazy boy [7]
3 years ago
12

Un convertidor catalítico acelera la reacción del monóxido de carbono con oxígeno para producir dióxido de carbono. ¿De qué mane

ra cada uno de los siguientes cambios afectaría la velocidad de reacción que se muestra aquí? Sustente 2CO (g) + O2(g) → 2CO2(g) a). Quitar el catalizador. B). Eliminar parte de O2(g)
Chemistry
1 answer:
vaieri [72.5K]3 years ago
5 0

Answer:

a. Disminuye.

b. Disminuye.

Explanation:

¡Hola!

En este caso, considerando que tenemos una reacción en la cual se podría asumir que tanto la concentración de monóxido de carbono y la de oxígeno contribuye a la velocidad de reacción, de acuerdo a la siguiente cinética elemental:

r=k[CO]^2[O_2]

Al aplicar los siguientes cambios, resultaría:

a. En este caso, bien sabemos que la velocidad de reacción es aumentada al adicionar un catalizador ya que la constante de velocidad aumenta al disminuir la energía de activación; no obstante, al retirar el catalizador, la energía de activación aumentaría y por consiguiente la velocidad de reacción disminuiría.

b. En este caso, dado que la relación entre el oxígeno y la rapidez de reacción es directamente proporcional, es possible para nosotros determinar que al eliminar parte del oxígeno, la concentración disminuye y por ende la velocidad de reacción también.

¡Saludos!

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Answer: Reaction A: pi + glucose ⇒ glucose-6-phosphate + H2O ΔG = 13.8 kJ/mol

Reaction B: pi + frutose-6-phosphate ⇒fructose-1,6-biphosphate + H2O ΔG = 16.3kJ/mol

Explanation: ΔG is the representation of the change in Gibbs Free Energy and relates enthalpy and entropy in a single value, which is:

ΔG = ΔH - TΔS

where:

ΔH is enthalpy

T is temperature

ΔS is entropy (measure of the )

It can also predict the direction of the reaction with the conditions of temperature and pressure being constant.

When the change is positive, the reaction is non-spontaneous, which means the reaction needs external energy to occur. If the change is negative, it is spontaneous, i.e., happens without external help.

Analyzing the reaction, we see that reaction A and B have a positive ΔG, while reaction C is negative, so the reaction that are unfavorable or nonspontaneous are <u>reactions A and B</u>.

6 0
3 years ago
Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric che
Helga [31]

Answer:

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

Explanation:

2NO_2(g)\rightleftharpoons N_2O_4(g)

Initially

3.0 atm                 0

At equilibrium

(3.0-2p)                 p

Equilibrium partial pressure of NO_2=2.1atm=3.0-2p

p = 0.45 atm

The value of equilibrium constant wil be given by :

K_p=\frac{p_{N_2O_4}}{(p_{NO_2})^2}=\frac{p}{(3.0-2p)^2}

K_p=\frac{0.45}{(2.1)^2}=0.10

After addition of 1.5 atm of nitrogen dioxide gas equilibrium reestablishes it self :

2NO_2(g)\rightleftharpoons N_2O_4(g)

After adding 1.5 atm of NO_2:

(2.1+1.5) atm                0.45 atm

At second equilibrium:'

(3.6-2P)                     (0.45+P)

The expression of equilibrium can be written as:

K_p=\frac{p'_{N_2O_4}}{(p'_{NO_2})^2}

0.10=\frac{(0.45+P)}{(3.6-2P)^2}

Solving for P:

P = 0.37 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time:

= (0.45+P) atm = (0.45 + 0.37 )atm = 0.82 atm

Partial pressure of dinitrogen tetroxide after equilibrium is reached the second time is 0.82 atm.

5 0
3 years ago
2.50g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 12
Art [367]

Answer:

                      Molecular Formula  =  C₁₀H₈

Explanation:

Step 1: <u>Calculating Moles for each Element:</u>

C  =  Mass of CO₂ × (1 mol of CO₂÷ M.Mass of CO₂) × (1 mol of C ÷ 1 mol of CO₂)

C  =  8.60 g × (1 mol of CO₂÷ 44.011 g.mol⁻¹) × (1 mol of C ÷ 1 mol of CO₂)

C =  0.1954 mol

H  =  Mass of H₂O × (1 mol of H₂O ÷ M.Mass of H₂O) × (2 mol of H ÷ 1 mol of H₂O)

H  = 1.41 g × (1 mol of H₂O ÷ 18.106 g.mol⁻¹) × (2 mol of H ÷ 1 mol of H₂O)

H =  0.1557 mol

Step 2: <u>Calculate the Smallest whole number ratio as,</u>

                                 C                                                        H

                              0.1954                                               0.1557

                      0.1954/0.1557                                  0.1557/0.1557

                               1.25                                                       1

Multiply ratio by 4,

                                 5                                                         4

Result:

          Empirical Formula of Propane is C₅H₄

Step 3: <u>Calculate Molecular Formula:</u>

Molecular formula is calculated by using following formula,

                    Molecular Formula  =  n × Empirical Formula  ---- (1)

Also, n is given as,

                     n  =  Molecular Weight / Empirical Formula Weight

Molecular Weight  =  128.0 g.mol⁻¹

Empirical Formula Weight  =  5 (C) + 4 (H) =  64 g.mol⁻¹

So,

                     n  =  128.0 g.mol⁻¹ ÷ 64 g.mol⁻¹

                     n  =  2

Putting Empirical Formula and value of "n" in equation 1,

                    Molecular Formula  = 2 × C₅H₄

                    Molecular Formula  =  C₁₀H₈

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3 years ago
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ANTONII [103]
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