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suter [353]
3 years ago
9

Name three high temperature reactions

Chemistry
1 answer:
lions [1.4K]3 years ago
3 0

Explanation:

As you increase the temperature the rate of reaction increases. As a rough approximation, for many reactions happening at around room temperature, the rate of reaction doubles for every 10°C rise in temperature. You have to be careful not to take this too literally. It doesn't apply to all reactions. Even where it is approximately true, it may be that the rate doubles every 9°C or 11°C or whatever. The number of degrees needed to double the rate will also change gradually as the temperature increases.

Examples

Some reactions are virtually instantaneous - for example, a precipitation reaction involving the coming together of ions in solution to make an insoluble solid, or the reaction between hydrogen ions from an acid and hydroxide ions from an alkali in solution. So heating one of these won't make any noticeable difference to the rate of the reaction. Almost any other reaction you care to name will happen faster if you heat it - either in the lab, or in industry.

The explanation

Increasing the collision frequency

Particles can only react when they collide. If you heat a substance, the particles move faster and so collide more frequently. That will speed up the rate of reaction. That seems a fairly straightforward explanation until you look at the numbers! It turns out that the frequency of two-particle collisions in gases is proportional to the square root of the kelvin temperature. If you increase the temperature from 293 K to 303 K (20°C to 30°C), you will increase the collision frequency by a factor of:

(1)303293=1.017" role="presentation" style="display: inline-table; font-style: normal; font-weight: normal; line-height: normal; font-size: 12px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; width: 10000em !important; position: relative;">303293=1.017(1)(1)303293=1.017

That's an increase of 1.7% for a 10° rise. The rate of reaction will probably have doubled for that increase in temperature - in other words, an increase of about 100%. The effect of increasing collision frequency on the rate of the reaction is very minor. The important effect is quite different . . .

The key importance of activation energy

Collisions only result in a reaction if the particles collide with enough energy to get the reaction started. This minimum energy required is called the activation energy for the reaction. You can mark the position of activation energy on a Maxwell-Boltzmann distribution to get a diagram like this:

Only those particles represented by the area to the right of the activation energy will react when they collide. The great majority don't have enough energy, and will simply bounce apart. To speed up the reaction, you need to increase the number of the very energetic particles - those with energies equal to or greater than the activation energy. Increasing the temperature has exactly that effect - it changes the shape of the graph. In the next diagram, the graph labeled T is at the original temperature. The graph labeled T+t is at a higher temperature.

If you now mark the position of the activation energy, you can see that although the curve hasn't moved very much overall, there has been such a large increase in the number of the very energetic particles that many more now collide with enough energy to react.

Remember that the area under a curve gives a count of the number of particles. On the last diagram, the area under the higher temperature curve to the right of the activation energy looks to have at least doubled - therefore at least doubling the rate of the reaction.

Summary

Increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions. It is only these collisions (possessing at least the activation energy for the reaction) which result in a reaction.

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A major component of gasoline is octane C8H18. When liquid octane is burned in air it reacts with oxygen O2 gas to produce carbo
Nady [450]

Answer:

0.054 moles of carbon dioxide will be produced

Explanation:

Step 1: Data given

Number of moles oxygen = 0.085 moles

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles of carbon dioxide

For 2 moles octane we need 25 moles oxygen gas to produce 16 moles carbon dioxide gas and 18 moles water vapor

0.0544

For 0.085 moles O2 we'll have 16/25 * 0.085 = 0.054 moles CO2

0.054 moles of carbon dioxide will be produced

7 0
3 years ago
helium gas in a 2.00 L cylinder is under 1.12 atm pressure. At 36.5°C that same gas sample has a pressure of 2.56 atm. What was
Irina18 [472]

Answer:

T₁ = 135.41 K

Explanation:

Given data:

Initial pressure = 1.12 atm

Finial temperature = 36.5 °C (36.5 +273 = 309.5 K)

Initial temperature = ?

Final pressure = 2.56 atm

Formula:  

P₁/T₁ = P₂/T₂  

P₁ = Initial pressure

T₁ = Initial temperature

P₂ = Final pressure

T₂ = Final temperature

Solution:

P₁/T₁ = P₂/T₂  

T₁  = P₁T₂  /P₂

T₁ = 1.12 atm × 309.5 K / 2.56 atm

T₁ = 346.64 atm . K / 2.56 atm

T₁ = 135.41 K

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posledela

Answer:

p=\frac{nRT}{V}

Explanation:

The ideal gas law equation is an equation that relates some of the quantities that describe a gas: pressure, volume and temperature.

The equation is:

pV=nRT

where

p is the pressure of the gas

V is the volume of the gas

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas (must be expressed in Kelvin)

Here we want to solve the equation isolating p, the pressure of the gas.

We can do that simply by dividing both terms by the volume, V. We find:

p=\frac{nRT}{V}

So, we see that:

- The pressure is directly proportional to the temperature of the gas

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