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4200 J of heat are added to a system. It does 5900 J of work on the surroundings. What is ΔE in J?

If the correct answer is positive, it means the system gained energy. If it is negative, then the system lost energy.

+4200J -5900J = -1700 J

3500 J of heat are lost from a gas to cooler surroundings. At the same time the gas is compressed and the surroundings do 3100 J of work on the gas. What is ΔE for the gas in J?

If the correct answer is positive, it means the system gained energy. If it is negative, then the system lost energy.

-3500J +3100J = -400J

A gas is compressed from 85.8 L to 1.6 L at 0.50 atmospheres? In addition a total of 5700 J of heat are gained from warmer surroundings. What is the change in internal energy in **kJ**?

If the correct answer is positive, it means the system gained energy. If it is negative, then the system lost energy.

w= -P delta V

-(0.5)(1.6-85.8) * 101.33= 4265.993

+5700 + 4265.993 = 9965.993/1000 = 9.965 kJ

The combustion of 1.00 mol of liquid methly alcohol (CH_{3}OH) at constant pressure in excess oxygen is exothermic, giving 727 kJ of heat per mole of CH_{3}OH. Write the balanced combustion reaction using smallest whole numbers. What is the coefficient for oxygen?

(CH_{3}OH + 3/2 O2 --> CO2 + 2H2O)2 dH= -727 kJ/mol

2CH_{3}OH + **3 O2 **--> 2CO2 + 4H2O

What is the enthalpy change (in kJ) for burning 16.0 g of methyl alcohol?

Hint given in feedback

Hint given in feedback

What is the sign of the change in enthalpy for an exothermic reaction? Hint, when a reaction is exothermic,what is the sign (positive or negative) for q_{surr}? Remember, energy is conserved. So, ΔH +q_{surr} = 0. In other words when one is positive the other is negative.)

Remember that the 727 kJ is per mole of CH_{3}OH.

16 g CH3OH * 1mol/ 3Lg = 0.5 mol CH3OH * -727kJ/mol = -363.5 kJ/mol

A reaction that is used to propel rockets is *N*_{2}O_{4}(l) + 2N_{2}H_{4}(l) --> 3N_{2}(g) + 4H_{2}O(g). This reaction has the advantage that neither product is toxic, so no dangerous pollution is released. When the reaction consumes 10.0 g of liquid N_{2}O_{4}, it releases 124 kJ of heat. What is the value of ΔH (in kJ per mole N_{2}O_{4}) for the chemical equation as written?

*N _{2}O_{4}(l) + 2N_{2}H_{4}(l) --> 3N_{2}(g) + 4H_{2}O(g)*

*-124 kJ/ 10g = -12.4 kJ/g * 92g/mol = -1140.8 kJ/mol*

Ammonia is produced commercially by the direct reaction of the elements. The formation of 1.00 moles of gaseous NH_{3} by this reaction releases 46.17 kJ of heat. How much energy (in kJ) is released, when 43.0 kg of hydrogen gas, H_{2}, reacts in excess nitrogen gas, N_{2}?

Hint given in feedback

Hint given in feedback

3H2+N2--> 2NH3

43 kg *1000g/1 kg =43000g H2* 1 mol/Lg= 21500 mol H2 * 2molH3/3molH2

=14333.3molNH3*46.17kJ/1mol=661770kJ/mol

The thermochemical equation for the burning of ethyl alcohol is

*C*_{2}H_{5}OH(l) + 3O_{2}(g) --> 2CO_{2}(g) + 3H_{2}O(l) ΔH = -1,367 kJ

What is the enthalpy change (in kJ) for burning 6.65 g of ethyl alcohol?

What is the enthalpy change (in kJ) for burning 6.65 g of ethyl alcohol?

C2H5OH + 3O2-->2CO2 + 3H2O dH=-1367 kJ

6.65g C2H5OH * 1 mol/ 46 g/mol = 0.145 mol C2H5OH*-1367kJ/mol

=-197.62 kJ/mol

A 85.1 g sample of metal at 80.91 ^{o}C is added to 48.84 g of water that is initially at 22.51 ^{o}C. The final temperature of both the water and the metal is 30.09 ^{o}C. The specific heat of water is 4.184 J/(g^{o}C). Calculate the specific heat of the metal.

85.1g(x)(30.09-80.91)+ 48.84g(4.18)(30.09-22.51)=0

x=0.358 J/g C

A 52.5 g sample of Al at 93.7 ^{o}C is added to 189.6 g of water that is initially at 24.2 ^{o}C. The specific heat of Al and water are 0.90 J/(g^{o}C) and 4.184 J/(g^{o}C), respectively. Calculate the equilibrium temperature.

(52.5g)(0.9)(x-93.7)+(189.6g)(4.18)(x-24.2)=0

x=28.11 C

A 65 mL solution of a dilute AgNO_{3} solution is added to 60 mL of a base solution in a coffee-cup calorimeter. As AgOH (s) precipitates, the temperature of the solution increases from 23.55 ^{o}C to 25.60 ^{o}C. Assuming the mixture has the same specific heat (4.184J/g^{o}C) and density (1.00 g/cm^{3} or 1.00 g/mL^{}) as water, calculate the heat (in J) transferred to the surroundings, q_{surr}.

Hint given in feedback

Hint given in feedback

1 mL = 1 cm^{3}. Now use the density of water to go from cm^{3} to g.

q(surr) = -q(sys)

(125g)(4.18)(25.6-23.55)=q

1071.125=q(surr)

Is the precipitation reaction exothermic or endothermic?

Select one:

The reaction(system) is releasing heat so the solution (surrounding) is getting warmer. Therefore the reaction is an exothermic reaction.

A 161 mL solution of a dilute acid is added to 120 mL of a base solution in a coffee-cup calorimeter. The temperature of the solution increases from 23.45^{o}C to 27.07 ^{o}C. Assuming the mixture has the same specific heat (4.184J/g^{o}C) and density (1.00 g/cm^{3}) as water, calculate the heat (in J) transferred to the surroundings, q_{surr}.

Hint given in feedback

Hint given in feedback

1 mL = 1 cm^{3}. Now use the density of water to go from cm^{3} to g.

(281g)(4.184)(27.07-23.45)=q

1174.58(3.62)=q

4251.98=q

Is the neutralization reaction exothermic or endothermic?

Select one:

Dissolving 6.51 g of CaCl_{2} in enough water to make 345 mL of solution causes the temperature of the solution to increase by 3.51 ^{o}C. Assume the specific heat of the solution and density of the solution are the same as water′s (about 4.18 J/g^{o}C and 1.00 g/cm^{3}, respectively) Calculate ΔH per mole of CaCl_{2} (in kJ) for the reaction under the above conditions.

Hint given in feedback

Aside, the ΔH per mole for dilution depends on the process. For example, more energy is released when starting with a large volume of water (infinite dilution), than when starting with a small volume of water. Why do you think this happens?

Hint given in feedback

Aside, the ΔH per mole for dilution depends on the process. For example, more energy is released when starting with a large volume of water (infinite dilution), than when starting with a small volume of water. Why do you think this happens?

Find the heat released per gram and convert to per mole using the MM. Is the process exothermic? If, yes, then what is the sign for ΔH?

(345g)(4.18)(3.51 C)=q

-5061.771J/0.058 mol CaCl2 = 87271.91 J/mol * 1kJ/1000J= -87.27 kJ

In the process of isolating iron from its ores, carbon monoxide reacts with iron (III) oxide, as described by the equation:

*Fe*_{2}O_{3}(s) + 3CO(g) --> 2Fe(s) + 3CO_{2}(g) ΔH = -24.8 kJ

The enthalpy change for the combustion of carbon monoxide is:

*2CO(g) + O*_{2}(g) --> 2CO_{2}(g) ΔH = -566 kJ

Use the preceding thermochemical equations to calculate the enthalpy change (in kJ) for the equation:

*4Fe(s) + 3O*_{2}(g) --> 2Fe_{2}O_{3}(s)(Use 4 sig. fig.)

The enthalpy change for the combustion of carbon monoxide is:

Use the preceding thermochemical equations to calculate the enthalpy change (in kJ) for the equation:

*-2(Fe _{2}O_{3}(s) + 3CO(g) --> 2Fe(s) + 3CO_{2}(g) ΔH = -24.8 kJ)*

**4Fe(s)** + ~~6CO~~_{2}(g)*--> 2Fe_{2}O_{3}(s) + *

*3(2CO(g) + O _{2}(g) --> 2CO_{2}(g) ΔH = -566 kJ)*

~~6CO(g)~~ + **3O _{2}(g)** -->

*4Fe(s) + 3O _{2}(g) --> 2Fe_{2}O_{3}(s) ΔH= 49.6-1698= -1648 kJ*

At 25^{o}C, the following heats of reaction are known:

2C_{2}H_{2}(g)+5O_{2}(g) → 4CO_{2}(g) + 2H_{2}O(l) ΔH = -2,600 kJ

C(s) + O_{2}(g) → CO_{2}(g) ΔH = -394 kJ

2H_{2}(g) + O_{2}(g) → 2H_{2}O(l) ΔH = -572 kJ

At the same temperature, calculate ΔH for the reaction:

2C(s) + H_{2}(g) → C_{2}H_{2}(g)

-1(2C_{2}H_{2}(g)+5O_{2}(g) → 4CO_{2}(g) + 2H_{2}O(l) ΔH = -2,600 kJ)

~~4CO~~(g) + _{2}~~2H~~(l)→ _{2}O**2C _{2}H_{2}(g)**+

4(C(s) + O_{2}(g) → CO_{2}(g) ΔH = -394 kJ)

**4C(s)** + ~~4O~~ → _{2}(g)~~4CO~~(g)_{2}

**2H _{2}(g)** +

4C(s) + 2H_{2}(g) → 2C_{2}H_{2}(g)

2C(s) + H_{2}(g) → C_{2}H_{2}(g)

(-1(2600) + 4(-394) -572) /2

226kJ

Use the standard enthalpies of formation from Appendix II (A-7 to A-12) to calculate the enthalpy change (in kJ) for the photosynthesis of glucose at 298 K and 1 atm.

*6CO*_{2}(g) + 6H_{2}O(l) --> C_{6}H_{12}O_{6}(s) + 6O_{2}(g)

Use 4 SF.

Use 4 SF.

Now Heat of formation of CO2 = -394 kJ

Heat of formation of H2O = -286

Heat of formation of Glucose = -1271 kJ

So Heat of Formation of Reaction = -1271 - 6 * (-394) - 6 * (286) = 2801J

Is the preceding reaction endothermic or exothermic?

Select one:

The reaction is endothermic because the heat of formation is positive

Use the standard enthalpies of formation from Appendix II (A-7 to A-12) to calculate the enthalpy change (in kJ) for the reduction of iron (III) oixde to iron at 298 K and 1 atm. (Calculate it for the reaction as written, namely 2 moles of iron (III) oxide and 3 moles of carbon.)*2Fe _{2}O_{3}(s) + 3C(s) --> 4Fe(s) + 3CO_{2}(g)*

Is the preceding reaction endothermic or exothermic?

Select one:

The reaction is endothermic because the heat of formation is positive

Frederick Trouton noted that the enthalpy of vaporization of 1 mole of a pure liquid, expressed in joules, is approximately 88 times the boiling point of the liquid on the Kelvin scale. This relationship is called Trouton's rule and is represented by the thermochemical equation: liquid → gas, ΔH_{vap} = 88 T_{b}joules/mol.

Combined with an empirical formula from chemical analysis, Trouton's rule can be used to find the molecular formula of a compound, as illustrated here. A compound that contains only carbon and hydrogen is 85.6% C and 14.4% H. Its enthalpy of vaporization is 389 J/g, and it boils at a temperature of 322 K.

First, determine the empirical formula of this compound using the percent composition of H and C. What is the empirical molar mass (in g/mol)?

Combined with an empirical formula from chemical analysis, Trouton's rule can be used to find the molecular formula of a compound, as illustrated here. A compound that contains only carbon and hydrogen is 85.6% C and 14.4% H. Its enthalpy of vaporization is 389 J/g, and it boils at a temperature of 322 K.

First, determine the empirical formula of this compound using the percent composition of H and C. What is the empirical molar mass (in g/mol)?

Use Trouton's rule to calculate the approximate enthalpy of vaporization (in J) of one mole of the compound.

Using the result from the preceding question and the experimental ΔH_{vap} of 389 J/g, find the approximate molar mass (in g/mol).

Hint given in feedback

Hint given in feedback

This is a simple multiplication. Using units, you can see how to multiple the information to get the approximate molar mass or g/mol.

What is the molar mass correct to 4 SIG. FIG.?

Hint given in feedback

Hint given in feedback

You know the approximate molar mass. Use the empirical formula to get the actual molar mass.

The price of gold is $625 per troy ounce at this writing. How much heat (in J) is needed to raise the temperature of $5000.00 worth of gold from 14.2 ^{o}C to 80.6 ^{o}C?

(1 troy oz = 31.10 g and the specific heat of Au is 25.42 J/(mol^{o}C).)

Review of multiplication by 1. Note, periodic table has additional needed information.

(1 troy oz = 31.10 g and the specific heat of Au is 25.42 J/(mol

Review of multiplication by 1. Note, periodic table has additional needed information.

Problems27-31. The law of Dulong and Petit states that the approximate molar heat capacity of a metal is 25 J/(mol^{.}K).

Experimentally, the specific heat of a metal is found to be 0.460 J/(g^{o}C). Use the law of Dulong and Petit to calculate the approximate molar mass of the metal.

Hint given in feedback

Experimentally, the specific heat of a metal is found to be 0.460 J/(g

Hint given in feedback

A chloride of this element is 67.2% chlorine by mass. In a 100 g sample how many moles of chlorine are there? (This info will be used to determine the empirical formula. Because the accuracy of the result in the last question for this set of questions depends on the accuracy of this answer, use the atomic mass of Cl accurate to 4 SIG. FIG.).

In the same 100 g sample __approximately__ how many moles of metal are there? (This info will be used to determine the empirical formula.)

Hint given in feedback

Hint given in feedback

Hint, use the approximate molar mass found earlier.

Determine the subscripts x and y in the empirical formula of the chloride. M_{x}Cl_{y} (The mole ratio of the elements you find will not be exactly whole numbers, so considerable rounding is needed to obtain whole numbers in the formula.) What is y?

Answer must be a whole number.

You know how many g of metal are in the 100 g sample. You can find the actual (not just approximate) moles of metal because you know the moles of chlorine in a 100 g sample and you know the relationship between moles of metal and moles of Cl (preceding question). **Determine an accurate molar mass for the metal. **ANSWER NEEDS TO BE RIGHT FOR 3 SIG. FIG.

Note, your approximate molar mass for the metal will indicate whether you are doing the calculation correctly. By the way, what is the metal? (Don't include in answer.)

Note, your approximate molar mass for the metal will indicate whether you are doing the calculation correctly. By the way, what is the metal? (Don't include in answer.)