Vocabulary
Formulas
Practice problems
Practice Problems
Practice Problems
100

What does the first law of thermodynamics state?

Energy can not be created nor destroyed just transferred. 

100

Two formulas for finding work?

w=(f)(d) and w=(-P^V)

100


Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4 and react to form a solid product C(s): A1g2 + B1g2 ¡ C1s2. As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system?


∆E = q + w = 1-1150J2 + 1480J2 = -670J


100

(a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of
the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 °C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?


a) 4.9*10^5J

b)11K/C

100


Calculate ∆H for the reaction 

NO(g) + O(g) --> NO21(g)

given this following information

NOg + O3g -->NO2(g) + O2(g)   ∆H = -198.9kJ

O3g --> 3/2O2(g) ∆H=-142.3kJ 

O2(g) --> 2Og  ∆H = 495.0kJ


-304.1kj

200

Define potential and kinetic energy.

Potential Energy= Energy possessed due to position or chemical composition 

Kinetic Energy: Energy possessed due to position

200

What is the formula for finding internal energy and how do you know if energy was gained or lost?

E=q+w or E=Final energy - Initial Energy 

200

A fuel is burned in a cylinder equipped with a piston. The initial volume of the cylinder is 0.250 L, and the final volume is 0.980 L. If the piston expands against a constant pressure of 136.8 kPa, how much work (in J) is done?



w = -P∆V = -1136.8 kPa210.730 L2 = 99.8 L@kPa


200


When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 to 23.11 °C. The temperature increase is caused by the following reaction:

AgNO3(aq) + HCl(aq) --> AgCl(s) + HNO3(aq)

Calculate ∆H for this reaction in kJ/mol AgNO , assuming that the combined solution has a mass of 100.0g and a specific heat of 4.18 J/g-°C.


-68.kj/mol or -68,000J/mol

200

During a  deep breath, our lungs expand about 2.0L against an external pressure of 101.3 kPa. How much work is in- volved in this process (in J)?


w=203 j

300

Meaning of exothermic and endothermic. 

Exothermic: Energy is being released 

Endothermic: Energy is being absorbed

300

This formula is used to find what? 

Eel=k(Q1Q2)/d

Electrostatic Potential Energy which shows the interaction between two charged particles. 

300


How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system?


-250kJ


300

When a student mixes 50 mL of 1.0M HCl and 50mL of 1.0M NaOH in a coffee cup calorimeter, the temperature of the resultant solution increases from 21.0 to 27.5 °C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g-K.


∆H = -2.7 kJ/0.050 mol = -54 kJ/mol



300

(a)What amount of heat(injoules)is required to raise the temperature of 1 g of water by 1 kelvin? (b) What amount of heat (in joules) is required to raise the temperature of 1 mole of water by 1 kelvin? (c) What is the heat capacity of 370 g of liquid water? (d) How many kJ of heat are needed to raise the temperature of 5.00 kg of liquid water from 24.6 to 46.2 °C?


a) 4.184J

b)75.4 J

c)1548 J/k

d) 452 kj

400

Difference between internal energy and enthalpy?

Internal energy is the sum of all kinetic and potential energies within the system. Enthalpy is the sum of heat within the system. 
400

Formula for enthalpy and change in enthalpy?

H=E+PV or change in ^H=Hproducts-Hreactants

400


Hydrogen peroxide can decompose to water and oxygen by the reaction

2H2O2(l)+ 2H2O(l) + O2(g) ∆H = -196kJ 

Calculate the quantity of heat released when 5.00g of H2O2(l) decomposes at constant pressure.

question is on pg:222

-14.4kj

400


The combustion of methylhydrazine CH6N2, a liquid rocket fuel, produces N2(g) + CO2(g), and H2O(l)

2CH6N2(l)+5O2(g) --> 2N2(g)+2CO2(g)+6H2O(l)

When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 to 39.50 °C. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794 kJ/°C. Calculate the heat of reaction for the combustion of a mole of CH6N2.


-1.30 * 10^3kJ/mol CH6N2


500

What is the relationship between the repulsion and attraction in relation to electro potential energy and charges of atoms (ie; like charges/opp. charges).

Like Charges: As the particles get closer, repulsion increases, Eel increases. As the particles move farther away, repulsion decreases, Eel decreases.

Opposite Charges: As the particles get closer, attraction increases, Eel decreases. As the particles move farther apart, attraction decreases, Eel increases. 

500

What is this formula used to find?

Cs=J/(g)(^T)

Specific Heat

500


a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room temperature) to 98 °C (near its boiling point)? (b) What is the molar heat capacity of water?



a) 7.9 * 10^4 J

b)75.2J/mol-K


500

Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is -393.5 kJ/mol, and that of diamond is -395.4 kJ/mol:

C1graphite + O2g --> CO2(g) ∆H = -393.5kJ C1diamond + O2g -->CO2(g) ∆H = -395.4kJ

Calculate ∆H for the conversion of graphite to diamond:

C1graphite2 --> C1diamond2 ∆H = ?


+1.9kJ