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Serjik [45]
4 years ago
13

What can be said about an endothermic reaction with a negative entropy change? View Available Hint(s) The reaction is What can b

e said about an endothermic reaction with a negative entropy change? spontaneous at all temperatures. spontaneous at high temperatures. spontaneous at low temperatures. spontaneous in the reverse direction at all temperatures. nonspontaneous in either direction at all temperatures. Submit Part B What can be said about an exothermic reaction with a negative entropy change? View Available Hint(s) The reaction is What can be said about an exothermic reaction with a negative entropy change? spontaneous at all temperatures. spontaneous at high temperatures. spontaneous at low temperatures. spontaneous in the reverse direction at all temperatures. nonspontaneous in either direction at all temperatures.
Chemistry
1 answer:
REY [17]4 years ago
5 0

Answer: The reaction is spontaneous at low temperatures

Explanation:

Let's first remember the definition of certain terms;

Enthalpy (H) in thermodynamics is defined as the heat content of a reaction. While Entropy (S) in thermodynamics is termed as the quantification of disorder or randomness of a reaction.

Gibb’s free energy change helps to determine the direction of the reaction.

Using the Gibbs free energy equation to solve this question.

∆G = ∆H - T∆S

Where;

∆G = Gibbs free energy

∆H = change in enthalpy

T= temperature of the reaction

∆S = change in entropy.

The question states that it is an exothermic reaction with negative entropy. This means that the change in both enthalpy and entropy will be negative. That is;

∆H = >0 ( it's positive)

∆S = < 0 (negative)

Let's remember that an exothermic reaction generally releases energy to it surroundings. This energy is usually released in the form of heat. Therefore, the change in enthalpy H of an exothermic reaction will always be negative. A negative change in entropy S indicates that there is a decrease in disorder, with respect to the reaction.

Using Gibb’s free energy equation at constant temperature and pressure, we have;

∆G = ∆H - T∆S

Now, the change in enthalpy and change in entropy can be written as follows;

∆H = >0 ( it's positive)

∆S = < 0 (negative)

Substitute these values in the above equation;

∆G = ∆H - T - (∆S)

∆G = ∆H + T∆S

According to the sign convention seen in the equation above,

Change in ∆G will be negative <0 when the value ∆H is greater than T∆S.

This change can occur only at low temperatures. Thus, this reaction is spontaneous at low temperatures.

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germanium -66 has a half-life of 2.5 hours. you have 35 grams of radioactive material. after 16 days, how much radioactive mater
Maru [420]

Hey there!

A half-life means after a certain amount of time, half of that substance will be gone/changed after that time.

How many half-lives are in 16 days?

Multiply 16 by 24 (to get the total hours) and divide that by 2.5.

That is 153.6 half-lives.

35(0.5)^(153.6) represents the equation for this.

Simplify this and get 8.88 x 10^(-45).

There will be 2.02 x 10^(-45)g of germanium-66 sample remaining after 16 days.

Hope this helps!

6 0
3 years ago
Which statement is true about the total mass of the reactants during a chemical change?
Mice21 [21]
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The number of ions produced when a substance dissolves is always equal to the number of Adams in a chemical formula. What is inc
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Question 3: Which state of matter does not freely flow?
vfiekz [6]

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Read 2 more answers
Photodissociation of ozone (O3) can be described as producing an oxygen atom with heat of formation 247.5 kJ/mol. However, in re
Lapatulllka [165]

Answer:

0.2193 μm

Explanation:

The reaction showing the Photodissociation of ozone (O3) is given below as:

           O₃         +        hv   -------------------------->      O₂         +       O⁺  

H°        (142.9)                                                         (0)                  (438kJ/mol).

The first thing to do here is to determine the change in the enthalpy of the total reaction, this can be done by subtracting the change in the enthalpy of the reactant from the change in enthalpy in the product. Hence, we have:

ΔH° = [438 kJ/mol + 247.5 kJ/mol] - (142.9) = 542.6 KJ/mol.

This value, that is 542.6 KJ/mol will then be used in the determination of the value for the maximum wavelength that could cause this photodissociation.

Therefore, the maximum wavelength could cause this photodissociation ≤ h × c/ E = [ 1.199 × 10⁻⁴]/ 542.6 = 2.193 × 10⁻⁷ =  0.2193 μm

7 0
3 years ago
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