Chapter 5 notes – Flashcards
Unlock all answers in this set
Unlock answers| what is energy? |
| the capacity to do work or transfer heat |
| what is kinetic energy? |
| the energy associated with an object by virtue of its motion |
| what is potential energy? |
| the energy an object has by virtue of its position in a field of force |
| what is the kinetic energy equation? |
| KE = 1/2mv2 |
| what is the jules equation? |
| J = 1kg m2/s2 |
| what is the calorie equation |
| cal = 4.184J |
| what is the dietary calorie equation? |
| Cal = 1000cal = 1kcal |
| the label on a cereal box indicates that one serving (with skim milk) provides 250Cal. what is this energy in kJ? |
| 250Cal X (1000cal/1Cal) X (4.184J/1cal) X (1kJ/1000J) = 1.0 X 103kJ |
A good pitcher can throw a baseball so that it travels between 60 and 80 miles per hour. A regulation baseball weighing 143g travels 75 miles per hour (33.5m/s). What is the kinetic energy of this baseball in joules? In calories?
K.E = 1/2mv2 1 cal = 4.184J |
(1/2)(.143kg)(33.52) = 80.2J (80.2/4.184) = 19.2cal |
| what is the definition of heat? |
| the energy that flows into or out of a system because of a difference in temperature between the system and its surroundings |
| what is a system? |
| the specific part of the universe that is of interest in the study |
| what is surroundings? |
| part of the universe that interacts with the system |
| what is the exothermic process? |
a chemical reaction in which heat is given off by the system to the suroundings. Heat exits the reaction. Surroundings warm up in temperature. 2H2(g) + O2(g) ----> 2H2O(l) + energy H2O(g) ----> H2O(l) + energy |
| what is the endothermic process? |
a chemical reaction in which heat is absorbed by the system form the surroundings. Heat enters the reaction. Surroundings cool down in temperature energy + 2HgO(s) ----> 2Hg(l) + O2(g) energy + H2O(s) ----> H2O(l) |
| what is the First Law Of Thermodynamics? |
energy can be converted from one form to another, but cannot be created or destroyed
ΔE = q + w |
| what is the equation for when gas expands against a constant external pressure? |
| w = -PΔV |
| the work done when a gas is compressed in a sylinder is 462J. during this process, there is a heat transfer of 128J from the gas to the surroundings. calculate the energy change, ΔE, for this process |
ΔE = q + w ΔE = (-128J) + (+462J) = 334J |
| a sample of nitrogen gas expands in volume from 1.6L to 5.4L at constant temperature. what is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7atm? |
w = -PΔV 101.3J - 1L atm a) w = 0J b) ΔV = 5.4L - 1.6L = 3.8L P = 3.7atm w = (-3.7atm)(3.8L) = -14.1L atm X (101.3J/1L atm) = - 1430J |
| what is enthalpy? |
| the heat absorbed or released under constant-pressure conditions |
| what is the enthalpy reaction equation? |
| ΔH = H products - H reactants |
| what is a thermochemical equation? |
one which shows the enthalpy change as well as the mass relationships N2(g) + 3H2(g) ----> 2NH3(g) ΔH = -91.8kJ |
| what happens when you reverse the reaction of a thermochemical equation? |
the sign of ΔH changes
CaCO3(s) --Δ--> CaO(s) + CO2(g) ΔH = +178kJ CaO(s) + CO2(g) --Δ--> CaCO3(s) ΔH = -178kJ |
2Ag2S(s) + 2H2O(l) --> 4Ag(s) + 2H2S(g) + O2(g) ΔH = +595.5kJ
Given the equation above, calculate ΔH for the following reaction:
Ag(s) + 1/2H2S(g) + 1/4O2(g) --> 1/2Ag2S(s) + 1/2H2O(l) ΔH = ? |
ΔH = (-595.5kJ/4) ΔH = -148.9kJ |
How much heat is evolved when 266g of white phosphorus (P4) burn in air?
P4(s) + 5O2(g) --> P4O10(s) ΔH = -3013kJ
(P4 = 123.9g/mol) |
P4(s) + 5O2(g) --> P4O10(s) + 3013kJ g P4 --> mol P4 --> kJ of heat
266g P4 x (1mol P4/123.9g) x (3013kJ/1mol P4) = 6470kJ |
How much heat could you obtain from 10.0g of methane, assuming you have an exccess of O2?
CH4(g) + 1O2(g) --> CO2(g) + 2H2O(l) ΔH = -890.3kJ
(CH4 = 16g/mol) |
g CH4 --> mol CH4 --> kJ heat 10g CH4 x (1mol CH4/16g) x (890.3kJ/1mol CH4) = 556kJ |
How much heat is evolved when 9.07 x 105g of NH3 is produced according to the following equation?
N2(g) + 3H2(g) --> 2NH3(g) ΔH = -91.8kJ |
g NH3 --> mol NH3 --> kJ heat 9.07 x 105g NH3 x (1mol NH3/17g) x (91.8kJ/2mol NH3) = 2.45 x 106kJ |
| What is the heat capacity of an object? |
| the amount of heat required to raise its temperature by on edegreee Celcius (or one Kelvin) |
How much heat is given off when an 869g iron bar cools from 94oC to 5oC?
(Fe = 0.444J/goC) |
q = C x m x ΔT ΔT = 5oC - 94oC = -89oC q = (0.444J/goC) x (869g) x (-89oC) q = -34000J or -34kJ |
How much energy must be transferred to raise the temperature of a cup of coffee (250mL) from 20.5oC (293.7K) to 95.6oC (368.8K)? Assume that water and coffee have the same density (1.00g/mL), and specific heat capacity (4.184J/g K).
(D = m/V) |
q = C x m x ΔT mass of coffee = 250mL x (1g/mL) = 250g ΔT = 368.8K - 293.7K = 75.1K q = (4.184J/gk) x (250g) x (75.1k) q = 79000J or 79kJ |
♥ Use energy transfer as a way to find the specific heat of a metal ♥ 55.0g Fe at 99.8oC ♥ Drop into 2225g water at 21.0oC ♥ Water and metal come to 23.1oC ♥ What is the specific heat capacity of the metal? |
heat lost by metal = heat gained by H2O -qmetal = qwater - (C) x m x ΔT = (C) x m x ΔT - (C)(55g)(-76.7oC) = (4.184J/goC)(225g)(21oC) C = .469J/goC |
| What is the Law of Conservation of Energy? |
| energy can neither be created nor destroyed. However energy can be transferred from one object to another, and it can assume different forms |
Suppose 0.562g of graphite is placed in a claorimeter with an excess of oxygen at 25.0oC and 1atm. Excess O2 ensures that all carbon burns to form CO2. The graphite is ignited, and it burns according to the equation: C(graphite) + O2(g) --> CO2(g) On reaction, the calorimeter temperature rises from 25.0oC to 25.89oC. The heat capacity of the calorimeter and its contents was determined in a separate experiment to be 20.7kJ/oC. What is the molar heat of combustion of carbon in kJ/mol? |
q = CΔT q = 20.7kJ/oC x (25.89oC - 25.0oC) q = 18.4kJ <-- this is for .52g of C - 18.4kJ .562g C x (1mol C/12.01g) = .0468 (-18.4kJ/.0468mol) = -3.93 x 102kJ/mol |
Suppose 33mL of 1.20M HCl is added to 42mL of a solution containing excess sodium hydroxide, NaOH, in a coffee-cup calorimeter. The solution temperature, originally 25.0oC, rises to 31.8oC. Calculate the molar heat of neutralization. HCl(aq)+ NaOH(aq) --> NaCl(aq) + H2O(l) For simplicity, assume the heat capacity and density of the final solution in the sup are those of water (4.184J/goC and 1.00g/mL, respectively). Also assume the total volume of the solution equals the sum of the volumes of HCl(aq) and NaOH(aq). |
qsolution = C x m x ΔT 33mL + 42mL = 75mL x (1g/1mL) = 75g q = (4.184J/goC) x (75g) x (31.8oC - 25.0oC) = 2133.84J qreaction = -2.134kJ number of mol = (1.20mol/1L) x (0.033L/1) = .0396mol = (-2.134kJ/.0396mol) = -53.9kJ/mol |
The standard heat of formation of nitric acid, HNO3(aq) is ΔHof = -207.4kJ/mol. Which equation is associated with this value?
a) H(g) = N(g) + O3(g) --> HNO3(aq) b) 1/2H2(g) + 1/2N2(g) + 3/2O2(g) --> HNO3(aq) c) HNO3(aq) --> 1/2H2(g) + 1/2N2(g) + 3/2O2(g) d) HNO3(aq) --> H(g) + N(g) + 3O(g) e) 1/2H2(g) + 1/2N2(g) + O2(g) --> HNO3(aq) |
| b) 1/2H2(g) + 1/2N2(g) + 3/2O2(g) --> HNO3(aq) |
| What is the standard enthalpy of reaction from standard enthalpies of formation equation? |
| ΔHorxn = ∑nΔHof(products) - ∑mΔHof(reactants) |
From the standard enthalpies of formation listed below, calculate the enthalpy change, ΔHo, for the following reaction: C2H4(g) + 3O2(g) --> 2CO2(g) + H2O(l)
Substance (ΔHof)kJ/mol C2H4(g) +52.3 CO2(g) -393.5 H2O(l) -285.9 |
ΔHorxn = ∑ΔHof(products) - ΔHof(reactants) ΔHo = [2mol (-393.5kJ/mol) + 2mol (-285.9kJ/mol)] - (52.3kJ/mol) ΔHo = -1411kJ |
| What is Hess's Law? |
| When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. |
Calculate the standard form of enthalpy of formation of CS2(l) given that: C(graphite) + O2(g) --> CO2(g) ΔHorxn = -393.5kJ S(rhombic) + O2(g) --> SO2(g) ΔHorxn = -296.1kJ CS2(l) + 3O2(g) --> CO2(g) + 2SO2(g) ΔHorxn = -1072kJ |
C(graphite) + 2S(rhombic) --> CS2(l) C(graphite) + O2(gas) --> CO2(g) ΔH = -393.5kJ 2S(rhombic) + 2O2(g) --> 2SO2(g) ΔH = 2x - 296.1kJ CO2(g) + 2SO2(g) --> CS2(l) + 3O2(g) ΔH = +1072kJ C(graphite) + 2S(rhombic) -->CS2(l) ΔH = 86.3kJ |
