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Helen [10]
3 years ago
12

An endergonic reaction with a Δh and Δs can be changed into an exergonic reaction.

Chemistry
1 answer:
zheka24 [161]3 years ago
4 0

Full question:

This question is incomplete, here it is completed:

An endergonic reaction with a ______ ∆H and a ______ ∆S can be changed into an exergonic reaction by decreasing the temperature.

Option A: negative, positive

Option B: negative, negative

Option C: positive, positive

Option D: positive, negative

Answer:

Option B: negative, negative

Explanation:

The change in free energy (ΔG) of a system for a constant-temperature process is

ΔG = ΔH - TΔS

free energy is the energy available to do work. Thus, if a particular reaction is accompanied by a release of usable energy (that is, <u>ΔG is negative</u><u>), it is said to be</u><u> exergonic</u>. On the other hand, if a reaction consumes energy (that is, <u>ΔG is positive</u><u>), it is said to be </u><u>endergonic</u>.

Looking at the equation, we can see that if ΔH is negative and ΔS is negative, then ΔG will be negative only when TΔS  is smaller in magnitude than ΔH. This condition is met when T is small.

ΔG = ΔH - TΔS

           -        -

This means that the reaction proceeds spontaneously at low temperatures. At high temperatures, the reverse reaction becomes spontaneous. An example of that would be the following reaction:

NH₃(g) + HCl(g) → NH₄Cl(s)

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7 0
1 year ago
What pressure (in kilopascals) is exerted by 4.20 moles of Xenon gas in a 15.0 L container at 280.0 K?
aniked [119]

Answer: the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa

Explanation:

The question requires us to calculate the pressure, in kPa, connsidering the following information:

<em>number of moles = n = 4.20mol</em>

<em>volume of gas = V = 15.0L</em>

<em>temperature of gas = T = 280.0 K</em>

We can use the equation of ideal gases to calculate the pressure of the gas, as shown by the rearranged equation below:

PV=nRT\rightarrow P=\frac{nRT}{V}

Since the volume was given in L and the question requires us to calculate the pressure in kPa, we can use R in units of L.Pa/K.mol:

<em>R = 8314.46 L.Pa/K.mol</em>

Applying the values given by the question to the rearranged equation above, we'll have:

\begin{gathered} P=\frac{nRT}{V} \\  \\ P=\frac{(4.20mol)\times(8314.46L.Pa/K.mol)\times(280.0K)}{(15.0L)}=652\times10^3Pa \end{gathered}

Therefore, the pressure exerted by the gas is 652 x 10^3 Pa, which corresponds to 652 kPa.

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1 year ago
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