All categories

3 years ago

According to collision theory, what three factors govern the effect of temperature on the rate of a chemical reaction?

Chemistry

1 answer:

anyanavicka [17]3 years ago

3 0

Answer:

Collision theory states that the rate constant for a chemical reaction is composed

of three factors, (1) the absolute number of collisions, Z, between molecules; (2) The

fraction of collisions, f, with an energy greater than the activation energy; and (3) the

fraction of molecules, p, in which the molecules are in the correct orientation to react.

k = Zfp Equation 1

The absolute number of collisions, Z, increases with temperature. However, it has

been shown that at 25o

C, the increase in the number of collisions accompanying a 10o

increase in temperature accounts for only about 2% of the increase in the reaction rate.

Similarly, while it is important that molecules be in the proper orientation to react when

they collide, molecular orientation is independent of temperature. Thus it follows that the

major factor controlling reaction rates is the fraction, f, of molecules in the reaction

mixture with an energy greater than the activation energy. This factor, f, depends on the

absolute temperature. It has been shown that f is related to Ea by the following equation

Explanation:

You might be interested in

Explain the difference between heat and temperature. Does 1 L of water at 658°F have more, less, or the same quantity of energy

pogonyaev

The answer is- The energy of 1 L water at temperature 347.78 °C have more energy as 1 L of water at temperature 65°C.

Heat is a type of energy that causes a person's body to feel hot or cold.

While the temperature of an object is a parameter that indicates how hot or cold the object is.

How is the temperature in degree Fahrenheit converted to degree celsius?

To convert the temperature in Fahrenheit to Celsius, subtract 32 and multiply by 5/9.

° $C =( ^0F -32) * \frac{5}{9}$

Now, heat is a form of energy that flows from hotter object to colder object and temperature indicates whether the object is hot or cold by measuring its average kinetic energy.
Now, the given temperature of 1 L water is 658 °F. This temperature in degree celsius is calculated as-

°C $= (658 F-32) *\frac{5}{9} = 347.78 \° C$

Now, higher the temperature, higher is the energy of water. Thus, the energy of 1 L water at 347.78 °C have more energy as 1 L of water at 65°C.

To learn more about heat and temperature, visit:

brainly.com/question/20038450

#SPJ4

3 0

2 years ago

A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$

Reduction half reaction (Cathode) : $Ag^{+}(aq)+e^-\rightarrow Ag(s)$

Thus the overall reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

7 0

3 years ago

What is the percent of Cr in Cr2O3

IgorLugansk [536]

The percentage of Chromium in Chromium Oxide is calculated as follow,

Step 1:
Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u

Step 2:
Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
= $\frac{103.98}{151.98}$ × 100

= 68.41 %
So, the %age composition of chromium in chromium oxide is 68.41 %.

5 0

3 years ago

Read 2 more answers

A solution has a pH of 2.5. Answer the following questions.

olga_2 [115]

PH + pOH = 14 ⇒ pOH = 14 - pH
pOH = 14 - 2.5
pOH = 11.5

[H⁺] = 10^(-pH) = 10^(-2.5)
[H⁺] = 0.003 M
[OH⁻] = 10^(-pOH) = 10^(-11.5) = 3 × 10⁻¹² M
[OH⁻] = 3 × 10⁻¹² M

pH = 2.5 implies one significant digit

6 0

3 years ago

What could J.J. Thomson conclude from his experiments?

KonstantinChe [14]

Answer:C, atoms must balance positive and negative particles.

Explanation: Got it correct on edgenuity.

6 0

3 years ago