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vfiekz [6]
3 years ago
11

g Tom's roommate Bob spilled crystal violet solution on his lab coat, and Tom offers to use some lye (sodium hydroxide solution)

to clean it off. Bob suggest diluting the lye with water, but Tom argues that the cleaning action will be faster if concentrated lye is used. Bob counters that the speed of cleaning doesn't depend on the concentration of the lye. Who is correct? Use your results to support your conclusion.
Chemistry
1 answer:
irina [24]3 years ago
4 0

Answer:

Tom is correct. The rate of reaction of Crystal violet and NaOH is first order with respect to NaOH, hence, a higher concentration of NaOH corresponds to a higher rate of reaction; a faster reaction.

This means the speed of cleaning depends on the concentration of the lye used.

Explanation:

The reaction between Crystal violet and NaOH, represented as

CV⁺ + OH⁻ → CVOH

It is a reaction that is know to turn the violet colour of the crystal violet colourless.

The rate of the reaction is also known to be second order; first order with respect to Crystal violet and first order with respect to NaOH.

This means that the rate of reaction is directly proportional to the concentration of NaOH provided all other parameters such as the rate constant and the concentration of Crystal violet are constant.

Hence, the reaction becomes faster with an increased concentration of NaOH.

So, Tom is right, concentrated lye solution would remove the stain faster.

Bob is wrong.

Hope this Helps!!!

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Calculate the ΔG°rxn using the following information at 298K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -2
Mkey [24]

Answer:

ΔG°rxn = +50.8 kJ/mol

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It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:

<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>

In the reaction:

2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)

ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)

ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}

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And S°:

S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)

ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)

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ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK

<em>ΔG°rxn = +50.8 kJ/mol</em>

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