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qwelly [4]
2 years ago
12

What is the relationship between the concentration and the rate of the reaction?

Chemistry
1 answer:
zalisa [80]2 years ago
3 0

Answer:

When the concentration of all the reactants increases, more molecules or ions interact to form new compounds, and the rate of reaction increases. When the concentration of a reactant decreases, there are fewer of that molecule or ion present, and the rate of reaction decreases.

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The stronger the intermolecular forces, the _______the energy needed to separate the molecules, the ____________the boiling poin
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higher, higher

Explanation:

It takes more energy to rip apart stronger bonds (that's mostly just common sense there). The boiling point increases because it would take more energy to get the molecules to go from a stuck together liquid, to separating in a gaseous form.

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How does distillation separate a solvent from a solute?
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Answer:

Distillable mixtures may contain a solid in a liquid. This method takes advantage of the boiling point of the substances.

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How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
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