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Alenkinab [10]
3 years ago
14

For which of the following reactions is S° > 0. Choose all that apply. N2(g) + 3H2(g) 2NH3(g) 2C2H6(g) + 7O2(g) 4CO2(g) + 6H2

O(g) NH4I(s) NH3(g) + HI(g) 2H2O(g) + 2SO2(g) 2H2S(g) + 3O2(g) 2NO(g) + 2H2(g) N2(g) + 2H2O(l) Submit Answer
Chemistry
1 answer:
Lostsunrise [7]3 years ago
5 0

Answer: 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

NH_4I(s)\rightarrow NH_3(g)+HI(g)

2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

Explanation:

Entropy is the measure of randomness or disorder of a system.

A system has positive value of entropy if the disorder increases and a system has negative value of entropy if the disorder decreases.

1. N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

As 4 moles of gaseous reactants are changing to 2 moles of gaseous products,  the randomness is decreasing and the entropy is negative

2. 2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)

As 9 moles of gaseous reactants are changing to 10 moles of gaseous products,  the randomness is increasing and the entropy is positive.

3. NH_4I(s)\rightarrow NH_3(g)+HI(g)

As 1 mole of solid reactants is changing to 2 moles of gaseous products, the randomness is increasing and the entropy is positive.

4. 2H_2O(g)+2SO_2(g)\rightarrow 2H_2S(g)+3O_2(g)

As 4 moles of gaseous reactants is changing to 5 moles of gaseous products, the randomness is increasing and the entropy is positive

5. 2NO(g)+2H_2(g)\rightarrow N_2(g)+2H_2O(l)

As 4 moles of gaseous reactants is changing to 1 moles of gaseous products, the randomness is decreasing and the entropy is negative.

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Answer:

The answer is decreased temperature and increased salinity

Explanation:

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Cold water is denser than hot water:

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Saltier water is much denser than water that has less salt:

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At  \fbox{\begin \\363 K \end{minispace}}  temperature, a sample of neon gas will occupy 50.00 \text{ m}^{3} volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

\fbox{ \begin \\ V \propto T \end{minispace}}

Or,

\frac{V}{T}=k

Here, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> is proportionality constant.

Given information:

The initial volume of neon gas is 40.81 \text{ m}^{3} .

The final volume of neon gas is  50.00 \text{ m}^{3}.

The initial temperature value is 23.5 \text{ } ^{\circ} \text{C} .

To calculate:

The final temperature

Given Condition:

  • The pressure is constant.
  • Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

Here,

  • V_{1}is the initial volume of the gas.
  • V_{2} is the final volume of the gas.
  • T_{1} is the initial temperature of the gas.
  • T_{2} is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}                                                                  …… (2)

Step 3: Convert the given temperature  from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

T(\text{K}) = T(^{\circ}\text{C}) + 273.15                                      …… (3)

Substitute 23.5\text{ }^{\circ} \text{C} for T(^{\circ}\text{C})  in equation (3) to convert temperature from degree Celsius to Kelvin.

T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}

Step 4: Substitute 40.81 \text{ m}^{3}  for V_{1} ,  50.00 \text{ m}^{3} for V_{2}  and  296.65 \text{ K} for T_{1}  in equation (2) and calculate the value of T_{2} .

T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}

Important note:

  • The temperature must be in Kelvin.
  • The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

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Explanation:

If we add half mole in 1L volume than molarity will obviously be 0.5 M.

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Half will be molecules of Ca and half will be of Cl2 for 0.5M.

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