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Rainbow [258]
3 years ago
6

Predict whether ΔS° is greater than, less than, or approximately zero for each of the following reactions, and explain your choi

ce. (a) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g) (b) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) (c) CaCO3(s) → CaO(s) + CO2(g)
Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:

Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

  • When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

So,

The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

And

If the particles are loosely held , the entropy will increase .  

  • If in a reaction , more number of gaseous atoms are present in the product side , entropy will increase , i.e. Δ°S > 0
  • When liquid is converted to solid entropy decreases,  

As the molecules in liquid state are loosely packed and has less force of attraction between the molecules, but as it is converted to solid, the force of attraction between the molecule increases and hence entropy decreases.

  • If in a reaction , less number of gaseous atoms are present in the product side , entropy will decrease , i.e. Δ°S < 0

From the question ,

( a )  NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

Gaseous atoms -

Reactant - 1 + 5 = 6

Product - 4 + 6 = 10 ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

( b ) CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

Gaseous atoms -

Reactant - 1 + 2 = 3

Product - 1 + 2 = 3 ,

Since ,

Both the side the value of gaseous atoms are , hence , Δ°S = 0 .

( c ) CaCO₃(s) → CaO(s) + CO₂(g)

Gaseous atoms -

Reactant = 0

Product - 0 + 1 = 1 ,

Since ,

Hence ,

More number of gaseous atoms are present in the product side , So ,

entropy will increase , i.e. Δ°S > 0

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A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide
Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe

Next, the total moles:

n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol

After that, since the process is isobaric, we can compute the work as:

W=P(V_2-V_1)

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3

Thereby, the magnitude and direction of work turn out:

W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0
3 years ago
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TiliK225 [7]
It should be option (A) because conduction is the transfer of heat energy from one place to another by vibrations/motion of the molecules. I hoped I helped.
5 0
3 years ago
3. Use the chart to answer the following question. Amanda found a stone
REY [17]

Answer:

Quartz

Explanation:

6 0
3 years ago
The ratio of x to y is 4 to 12. Fill in numbers to<br>form a table with the given ratio,​
murzikaleks [220]

Answer:

Explanation:

x                                     y

4                                     12

8                                     24

12                                   36

16                                   48

20                                  60

           

4 0
3 years ago
The combustion of 1 mole of CO according to the reaction CO(g) + ½O2(g) → CO2(g) + 67.6 kcal gives off how much heat?
Tomtit [17]
When the enthalpy value is given, we can calculate how much heat is use or produces in a given equation. 

67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction=  1 mol of CO (based on the coefficient)

so 1 mole of CO gives us 67.6 kCal of heat.

calculation:

1 mol CO\frac{67.5 kcal}{1 mol CO} = 67.5 kcal




5 0
3 years ago
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