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USPshnik [31]
2 years ago
10

What is the enthalpy of the overall chemical equation

Chemistry
2 answers:
Dimas [21]2 years ago
5 0

Recall that standard enthalpies of formation can be either positive or negative. The enthalpy of formation of carbon dioxide at 298.15K is ΔH f = -393.5 kJ/mol CO 2(g). Write the chemical equation for the formation of CO2

12345 [234]2 years ago
3 0

Answer:

The enthalpy of the overall chemical equation is -304.1 kJ

Explanation:

Enthalpy is an additive property. The overall chemical equation can be obtained by summing up all three elementary steps. So enthalpy of overall reaction is summation of enthalpy changes in each steps.

                    NO(g)+O_{3}(g)\rightarrow NO_{2}(g)+O_{2}(g)

                    \frac{3}{2}O_{2}(g)\rightarrow O_{3}(g)

                    O(g)\rightarrow \frac{1}{2}O_{2}(g)

                 ----------------------------------------------------------------------------------------

                     NO(g)+O(g)\rightarrow NO_{2}(g)

Enthalpy of overall reaction = \Delta H_{1}+\Delta H_{2}+\Delta H_{3}

                                               = (-198.9+142.3-247.5) kJ

                                                = -304.1 kJ

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
0.500 L of a gas is collected at 2911 MM and 0°C. What will the volume be at STP?
ioda

Answer:

V₂ =  1.92 L

Explanation:

Given data:

Initial volume = 0.500 L

Initial pressure =2911 mmHg (2911/760 = 3.83 atm)

Initial temperature = 0 °C (0 +273 = 273 K)

Final temperature = 273 K

Final volume = ?

Final pressure = 1 atm

Solution:

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

by putting values,

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm

V₂ = 522.795 atm .L. K / 273 K.atm

V₂ =  1.92 L

4 0
3 years ago
Why do you think it is advantageous to use inert electrodes in the electrolysis process? ​
olga nikolaevna [1]

Answer:

The advantage of this technique is that purified water as well as deposited metals can be re-used. It is necessary to use an inert electrode, such as platinum, because there is no metal present to conduct the electrons from the anode to the cathode.

5 0
2 years ago
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