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pychu [463]
3 years ago
13

If a reaction mixture contains only N2O and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until som

e NO forms in the mixture. What temperature is required to make the reaction spontaneous under standard conditions?

Chemistry
1 answer:
Vesna [10]3 years ago
3 0

Answer:

Temperature required = 923K

Explanation:

The question is incomplete as there are some details that has to be given. details like the values of the standard enthalpies and entropies of the reactants and product as this is needed to calculate the actual value of the standard enthalpies and standard entropies of the reaction. I was able to get those values from literature and then calculated what needs to be calculated.

From there, I was able to use the equation that shows the relationship between, gibb's free energy, enthalpy, entropy and temperature. The necessary mathematical manipulation were done and the values were plugged in to get the temperature required to make the reaction spontaneous.

A few notes on the Gibb's free energy.

The Gibb's free energy also referred to as the gibb's function represented with letter G. it is the amount of useful work obtained from a system at constant temperature and pressure. The standard gibb's free energy on the other hand is a state function represented as Delta-G, as it depends on the initial and final states of the system.

The spontaneity of a reaction is explained by the standard gibb's free energy.

  • If Delta-G = -ve ( the reaction is spontaneous)
  • if Delta -G = +ve ( the reaction is non-spontaneous)
  • if Delta-G = 0 ( the reaction is at equilibrium)

The step by step calculations is done as shown in the attachment.

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Hey there!

27 tells us the sum of protons and neutrons is 27.

Al tells us we have 13 protons.

3+ tells us that there are 3 less electrons than protons.

13 + n = 27

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13 - 3 = 10 electrons

27Al3+ has 13 protons, 14 neutrons, and 10 electrons.

Hope this helps!

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Read 2 more answers
The pKa of chloroacetic acid is 2.9. If an organic chemist has a 0.10 M aqueous solution of chloroacetic acid at pH 2.4, what pe
Trava [24]

Answer:

The answer is given below.

Explanation:

We will consider the acid as HA and will set up an ICE table with the equilibrium dissociation of α.

AT pH 2.4 the initial H+ concentration will be 3.98^10-3 M

                      HA →              H+         +      A-

Initial concentration:  0.1    →  3.98 ^10-3       +      0

equilibrium concentration:  0.1(1-α) →   3.98 * 10-3 + 0.1α              0.1α

pKa of chloroacetic acid is 2.9

-log(Ka) = 2.9

Ka = 1.26 * 10-3

From the equation, Ka = [H+] * [A-] / [HA]

1.26 * 10-3 = (3.98 * 10-3 + 0.1α )* 0.1α / 0.1(1-α)      

Since α<<1, we assume 1-α = 1

Solving the equation, we have: α = 0.094

Since this is the fraction of acid that has dissociated, we can say that % of base form = 100 * α= 9.4%

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3 years ago
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