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Radda [10]
3 years ago
13

You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the followin

g statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.
Chemistry
1 answer:
m_a_m_a [10]3 years ago
8 0

Answer: option (a) is the correct answer

Explanation:

The complete questions says;

You observe an exothermic gaseous reaction that is not spontaneous in forward direction at 1 atm and 298K. Which of the following statements about this reaction is true? a. This reaction will become spontaneous in forward direction at some temperature below 298K. b. This reaction will be spontaneous in forward direction at a higher pressure at 298K. c. This reaction will become spontaneous in forward direction at some temperature above 298K. d. This reaction is never spontaneous. e. The reverse reaction is always spontaneous.

The Answer:

(a). This reaction will become spontaneous in forward direction at some temperature below 298K

Explanation: First of all, we can acknowledge that the reaction seen here is an exothermic one, i.e energy is released in the process outwardly and as a result temperature is reduced during this process of energy loss.

Having understood that scenario, say we reduce it's temperature by ourself than forward reaction favors and after reaching at particular temperature, therefore we can confirm this to be a spontaneous reaction.

Let us use this to confirm what we have been saying.

Given;

ΔG = ΔH - TΔS

here ΔH is negative

it is non spontaneous, which means ΔG is positive so we continuously decraeses it's temperature than at a particular temperature.

The Entropy change becomes positive and reaction becomes spontaneous and ΔG become negative.

cheers i hope this helped !!

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skad [1K]

The density of the sample : 0.827 g/L

<h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m2, v= m³)

T = temperature, Kelvin  

n= 1 mol

MW Neon = 20,1797 g/mol

mass of Neon :

\tt mass=mol\times MW\\\\mass=1\times 20,1797 =20.1797~g

The density of the sample :

\tt \rho=\dfrac{m}{V}\\\\\rho=\dfrac{20,1797}{24.4}=0.827~g/L

or We can use the ideal gas formula ta find density :

\tt \rho=\dfrac{P\times MW}{RT}\\\\\rho=\dfrac{1\times 20.1797}{0.082\times 298}\\\\\rho=0.826~g/L

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3 years ago
If you stretch back a rubber band and release it, it shoots across the room. What type of energy conversion has occurred?
MAVERICK [17]
<span>Elastic to mechanical is the energy conversion that occurs. Elastic, the pulling back and stretching of the rubber band, mechanical is the release!</span>
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3 years ago
Calculate the amount of heat needed to boil 41.1 g of water (H2O), beginning from a temperature of 84.7 C . Be sure your answer
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Explanation:

We need to go through to stages to boil 41.1 g of water. We have to heat the sample of water from 84.7 °C to 100 °C (the boiling point) And then we have to provide enough heat to boil all the sample of water.

<em>a) Heating from 84.7 °C to 100 °C:</em>

This is calculated using the formula:

Q₁ = m * C * ΔT

Where Q₁ is the amount of heat, m is the mass of the sample, C is the specific heat of water and ΔT is the temperature change. We already know these values:

m = 41.1 g

C = 4.184 J/(g*°C)

ΔT = Tfinal - Tinitial = 100 °C - 84.7 °C

ΔT = 15.3 °C

Replacing these values we can get the amount of heat necessary for the first step:

Q₁ = m * C * ΔT

Q₁ = 41.1 g * 4.184 J/(g°C) * 15.3 °C

Q₁ = 2631 J

<em>b) Boiling 41.1 g of water:</em>

To find the amount of heat that we need to provide to the sample of water to completely boil it we can use this formula:

Q₂ = m * Cv

Where Cv is the latent heat of vaporization.

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Q₂ = m * Cv

Q₂ = 41.1 g * 2256 J/g

Q₂ = 92721 J

<em>c) Total amount of heat:</em>

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Qtotal = 2631 J + 92721 J

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Qtotal = 95.4 kJ

Answer: The amount of heat needed to boil the sample of water is 95.4 kJ or 95400 J.

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<span>Therefore, for the reaction, the amount of calcium oxide produced is 0.032 moles.</span>
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