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Helga [31]
4 years ago
8

while the substance is a liquid at what temperature do the molecules of the substance have the greatest average kinetic energy ​

Chemistry
1 answer:
dalvyx [7]4 years ago
4 0

The molecules of the liquid will be having the greatest average kinetic energy when the temperature of the liquid reaches its boiling point.

<u> Explanation: </u>

Average kinetic energy is the measure of kinetic energy exhibited by the molecules of any substance. It is directly proportionate to the temperature of the substance. According to kinetic molecular theory,

                           k=\frac{3 R T}{2 N_{A}}

As the molecules in solids are mostly closely packed, so they will attain the least kinetic energy of molecules in solid phase. Similarly, the molecules in liquid are slightly loosely packed and on increasing the temperature, the molecules will attain more energy to increase their neighbouring distances.

And when that temperature reaches the boiling point of that liquid substance, the molecules had attained the maximum amount of energy to increase its average kinetic energy to the maximum limit and transferring the liquid molecules to gaseous state.

As the molecules in gas phase are very loosely packed. The starting of the change in the state of molecules from liquid to gas phase on reaching the boiling point indicate the maximum average kinetic energy at that point.

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tatuchka [14]

Answer:

rate = k[A][B] where k = k₂K

Explanation:

Your mechanism is a slow step with a prior equilibrium:

\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}

(The arrow in Step 1 should be equilibrium arrows).

1. Write the rate equations:

-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]

2. Derive the rate law

Assume k₋₁ ≫ k₂.  

Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.  

In an equilibrium, the forward and reverse rates are equal:

k₁[A][B] = k₋₁[C]

[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)

rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]

The rate law is  

rate = k[A][B] where k = k₂K

5 0
4 years ago
How many grams of O2(g) are needed to completely burn 73.7 g of C3H8(g)?
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The balanced chemical reaction is:

C3H8 + 5O2 = 3CO2 + 4H2O

We are given the amount of C3H8 to be used for the reaction. This will be the starting point for the calculations.

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4 years ago
For the following reaction, 7.43 grams of methane (CH4) are allowed to react with 27.6 grams of carbon tetrachloride. methane (C
NikAS [45]

Answer:

30.4 grams of CH2Cl2 will be formed

Explanation:

Step 1: data given

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Step 2: The balanced equation

CH4 + CCl4 → 2CH2Cl2

Step 3: Calculate moles

Moles = mass / molar mass

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For 1 mol CH4 we need 1 mol CCl4 to produce 2 moles CH2Cl2

For 0.179 moles CCl4 we'll have 2*0.179 = 0.358 moles CH2Cl2

Step 6: Calculate mass CH2Cl2

Mass CH2Cl2 = 0.358 moles* 84.93 g/mol

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<h2>Answer:</h2>
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