Chem 115 POGIL Worksheet - Week 8 - Answers ...

嚜澧hem 115 POGIL Worksheet - Week 8 - Answers

Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Key Questions & Exercises

1. Calculate 忖Ho for the reaction,

C2H2(g) + H2(g) ‾ C2H4(g)

Given:

(a) 2 C2H2(g) + 5 O2(g) ‾ 4 CO2(g) + 2 H2O(l)

(b) C2H4(g) + 3 O2(g) ‾ 2 CO2(g) + 2 H2O(l)

(c) H2(g) + ? O2(g) ‾ H2O(l)

忖Ho = 每2599.2 kJ

忖Ho = 每1410.9 kJ

忖Ho = 每285.8 kJ

The (a) equation has C2H2(g) on the left, where we need it in the target equation, but its

coefficient is 2. If we take one half of equation (a) we will have the correct coefficient. The

忖Ho contribution will be (?)(每2599.2 kJ) = 每1299.6 kJ. Equation (b) has C2H2(g) with the

correct coefficient for our target equation, but we need it on the right, not the left as given.

Therefore, we will reverse it, making its 忖Ho contribution be +1410.9 kJ. Finally, if we take

equation (c) as given, it will introduce the H2(g) with the correct coefficient, and it will result

in cancellations of the CO2(g), H2O(l), and O2(g) introduced by our manipulations of the first

two equations, which are absent in the target equation. The following sum results:

(?a)

(每b)

(c)

C2H2(g) + 5/2 O2(g) ‾ 2 CO2(g) + H2O(l)

忖Ho = 每1299.6 kJ

2 CO2(g) + 2 H2O(l) ‾ C2H4(g) + 3 O2(g)

忖Ho = +1410.9 kJ

H2(g) + ? O2(g)

‾ H2O(l)

忖Ho = 每285.8 kJ

_________________________________________________________

C2H2(g) + H2(g)

‾ C2H4(g)

忖Ho = 每174.5 kJ

2. Write the balanced thermochemical equations that pertain to the given compounds and their

standard enthalpies of formation.

Compound

忖Hof (kJ/mol)

CCl4(g)

每106.7

C(s) + 2 Cl2(g) ‾ CCl4(l)

Fe2O3(s)

每822.16

2 Fe(s) + 3/2 O2(g) ‾ Fe2O3(s)

HNO3(g)

每134.3

1/2 H2(g) + 1/2 N2(g) + 3/2 O2(g) ‾ HNO3(g)

NaHCO3(s)

每947.7

Na(s) + 1/2 H2(g) + C(s) + 3/2 O2(g) ‾ NaHCO3(s)

Thermochemical Equation

3. i. Given:

(a)

(b)

N2O4(g) + 1/2 O2(g) ‾ N2O5(g)

HNO3(g) ‾ 1/2 N2O5(g) + 1/2 H2O(l)

忖Ho = +1.67 kJ

忖Ho = 每2.96 kJ

Calculate 忖Ho for the reaction

N2O4(g) + H2O(l) + 1/2 O2(g) ‾ 2 HNO3(g)

忖Ho = ?

The two given equations, manipulated as shown below, will add to give the target

reaction and its enthalpy:

(a)

N2O4(g) + 1/2 O2(g)

‾ N2O5(g)

忖Ho = +1.67 kJ

‾ 2 HNO3(g)

忖Ho = +5.92 kJ

(每2b) N2O5(g) + H2O(l)

__________________________________________________________

N2O4(g) + H2O(l) + 1/2 O2(g) ‾ 2 HNO3(g)

忖Ho = 7.59 kJ

ii. Given following standard enthalpy of formation data:

Compound

忖Hof (kJ/mol)

N2O4(g)

+9.66

HNO3(g)

每134.31

H2O(l)

每285.83

Calculate 忖Ho for the reaction

N2O4(g) + H2O(l) + 1/2 O2(g) ‾ 2 HNO3(g)

忖Ho = ?

Compare your answer to your answer in part i.

The enthalpy for the reaction is the sum of the enthalpies of formation for products minus

those for reactants, each multiplied by its stoichiometric coefficient in the balanced

equation:

忖Ho = 2忖Hof (HNO3) 每 {忖Hof (N2O4) + 忖Hof (H2O) + ?忖Hof (O2)}

= (2)(每134.31 kJ) 每 {+9.66 kJ + (每285.83 kJ) + 0 kJ}

= 7.55 kJ

Note that the value of 忖Hof(O2) is taken as 0, because O2(g) is the standard state for

elemental oxygen.

As expected by the Law of Hess, the answers are virtually the same. In many cases there

is a slight difference, due to minor differences in the values from the experimental data.

4. Calculate the heat of combustion of methane, CH4(g), defined by the following

thermochemical equation:

CH4(g) + 2 O2(g) 6 CO2(g) + 2 H2O(l)

忖Horxn = ?

Given the following standard enthalpies of formation:

每74.85 kJ

CH4(g)

CO2(g)

每393.5 kJ

H2O(l)

每285.8 kJ

This can be calculated as follows:

忖Horxn = {忖Hof(CO2) + 2忖Hof(H2O)} 每 {忖Hof(CH4) + 2忖Hof(O2)}

= {(每393.5 kJ) + (2)(每285.8 kJ)} 每 {(每74.85 kJ) + 0}

= 每890.25 kJ = 每890.2 kJ

Note that the value of 忖Hof(O2) is taken as 0, because O2(g) is the standard state for elemental

oxygen.

5. Calculate the enthalpy of formation, 忖Hof, for benzene, C6H6(l), given that the heats of

formation of CO2(g) and H2O(l) are 每393.5 kJ and 每285.8 kJ, respectively, and that the heat

of combustion of C6H6(l) is 每3267.7 kJ. To do this, carry out the following steps.

i. Write the balanced thermochemical equation that defines the enthalpy of formation of

benzene, C6H6(l).

忖Hof(C6H6) = ?

6 C(graphite) + 3 H2(g) ‾ C6H6(l)

ii. Write the balanced thermochemical equation for the heat of combustion of benzene,

C6H6(l).

忖Hocomb = 每3267.7 kJ

C6H6(l) + 15/2 O2(g) ‾ 6 CO2(g) + 3 H2O(l)

iii. Based on your answer to question ii, write an expression for the heat of combustion of

benzene, 忖Hocomb, in terms of the enthalpies of formation of the reactants and products.

Using the data given in the problem, solve this for the unknown value of 忖Hof(C6H6), the

enthalpy of formation of benzene.

We can express the heat of combustion as

忖Hocomb = 6忖Hof(CO2) + 3忖Hof(H2O) 每 忖Hof(C6H6)

We have values for everything in this equation except 忖Hof(C6H6). Therefore, we can

rearrange and solve for it:

忖Hof(C6H6) = 6忖Hof(CO2) + 3忖Hof(H2O) 每 忖Hocomb

= (6)(每393.5 kJ) + (3)(每285.8 kJ) 每 (每3267.7 kJ) = +49.3 kJ

6. For each of the following, indicate which kind of radiation has higher energy.

red light or blue light

infrared radiation or radio waves

x-rays or visible light

7. For each of the following, indicate which has higher frequency

light with 竹 = 490 nm or light with 竹 = 520 nm (糸 = c/竹)

light with energy of 3.0 x 10每19 J or light with energy of 4.5 x 10每19 J (E = h糸)

8. An argon laser emits green light with a wavelength of 514.5 nm. Calculate the following for

this light: (a) the wavelength in ?; (b) the frequency in Hz (s-1); (c) the energy in joules, J.

(a)

Note that to convert from nanometers to ?ngstroms we need only move the

decimal point to the right one place.

(b) Use 糸 = c/竹. We need wavelength in meters, because the units on the speed of light are

mAs-1. We could convert nm to m first, but with a calculator it is simpler to enter

nanometers as 10每9 m; i.e., ※514.5E-9".

(c) Given the preceding results, we can calculate energy by either E = h糸 or E = hc/竹. Using

the first of these,

E = h糸 = (6.626 x 10每34 JAs)(5.827 x 1014 s每1) = 3.861 x 10每19 J

9. Is energy emitted or absorbed when the following transitions occur in hydrogen:

(a) from n = 1 to n = 3

The transition is to a higher energy state, so energy corresponding to the difference

between the states must be absorbed.

(b) from n = 5 to n = 2

The transition is to a lower energy state, so energy corresponding to the difference

between the states must be emitted.

(c) an H+ ion acquires an electron into the n = 2 state.

An H+ ion is essentially a hydrogen atom with an electron an infinite distance away, the

state for which n = 4. By adding an electron to the n = 2 state, it has lower energy (the

electron is closer to the nucleus to which it is attracted), so energy will be emitted

10. The four visible lines of the Balmer series in the emission spectrum of hydrogen are violet

(410 nm), blue (434 nm), blue-green (486 nm), and red (656 nm). Assign these as state-tostate transitions of the type ni ‾ nf, giving the n values involved for each line.

All transitions in the Balmer series are to nf = 2 from ni > 2. The energy of light increases

going from red to violet, so the red line is the smallest state-to-state separation, and the violet

line is the largest.

red (656 nm)

3‾2

blue-green (486 nm)

4‾2

blue (434 nm)

5‾2

violet (410 nm)

6‾2

11. Calculate the energy of the first line in the Lyman series for the hydrogen atom, which arises

from a transition from ni = 2 to nf = 1. What is the wavelength of the radiation emitted? In

what region of the electromagnetic spectrum does it fall?

Ephoton = 1.635 ℅ 10每18 J

A wavelength of 121 nm falls well within the ultraviolet region.

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