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Georgia [21]
2 years ago
15

Given the reactions below, (1) Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g) LaTeX: \Delta H^o_{rxn}Δ H r x n o = −146 kJ (2) Na2SO4(s) +

H2O(l) → 2NaOH(s) + SO3(g) LaTeX: \Delta H^o_{rxn}Δ H r x n o = 418 kJ (3) 2Na2O(s) + 2H2(g) → 4Na(s) + 2H2O(l) LaTeX: \Delta H^o_{rxn}Δ H r x n o = 259 kJ Determine LaTeX: \Delta H^o_{rxn}Δ H r x n o for the following reaction: (4) Na2O(s) + SO3(g) → Na2SO4(s) LaTeX: \Delta H^o_{rxn}Δ H r x n o = ?
Chemistry
1 answer:
slamgirl [31]2 years ago
4 0

Answer:

   ΔrxnH  = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work  in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking  1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

      Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l)             ΔrxnHº = 259 / 2 kJ  1/2 eq3

+    2NaOH(s)  + SO₃(g) ⇒  Na₂SO₄ (s) + H₂O (l)   ΔrxnHº = -418 kJ     - eq 2

+    2Na (s) + 2 H₂O (l)  ⇒   2 NaOH (s) + H₂ (g)    ΔrxnHº = -146 x 2    2 eq 1

<u>                                                                                                                                         </u>

Na₂O (s) + SO₃ (g)  ⇒ Na₂SO₄ (s)    ΔrxnHº =  259/2 + (-418) + (-146) x 2 kJ

                                                          ΔrxnH  = -580.5 kJ

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
3 years ago
Heat is transferred directly from a heat reservoir at 200 C to another heat reservoir at 5C. If the amount of heat transferred i
Daniel [21]

Answer:

ΔS=0.148  KJ/K

Explanation:

Given that

Q = 100 KJ

T₁=200°C

T₁=200+273 = 437 K

T₂=5°C

T₂=5 + 273 = 278 K

Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.

So the total change in entropy given as

ΔS=  - Q/T₁ + Q/T₂

ΔS=  - 100/473 + 100/278  KJ/K

ΔS=0.148  KJ/K

4 0
3 years ago
Does the volume of particles affect the behavior of gas
lilavasa [31]

Answer:

Yes, it does, although only physically and not chemically.

Explanation:

If a volume of gas is way spread out, it won't collide with the other gas particles as often, reducing pressure and temperature because they lose kinetic energy to their surroundings when they don't collide.

If it is compressed, it increases temperature and pressure because the gas particles collide with each other and the walls of the container way more often than if they had more space.

Hope this answers your question.

P.S.

Fun fact, gas particles are actually moving at 300-400 meters per second at room temperature, they only slow down to walking speed at very low temperatures, like 10 Kelvin

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Answer:

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