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fomenos
3 years ago
5

When ammonium chloride crystals are dissolved in water, the temperature of the water decreases. What does the temperature change

indicate about the dissolving of ammonium chloride in water.
Chemistry
1 answer:
sertanlavr [38]3 years ago
3 0

Answer:

See Explanation

Explanation:

A decreasing temperature indicates that the dissolution process for the ammonium chloride requires input of energy from surroundings. That is, the process is essentially 2 parts => system (object of interest - NH₄Cl) and the surroundings (everything else - solvent - H₂O). The surroundings (water) solvent is showing a <u>measured</u> decrease in temperature or loss of energy (exothermic to surroundings) which flows into the system (NH₄Cl) and effects dissolution of salt into solution (endothermic to system).

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What is the Law of Definite Composition and how does it apply to pure substances and mixtures?
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The reaction that has the greatest tendency to be reversed in an spontaneous redox reaction is that whose forward standard reduction potential is the lowest (mos negative) one because that means that the reversed reaction will have the highest (most positive) standard reduction potential.

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3 years ago
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

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A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

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