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

a chemical reaction in which heat is given off by the system to the suroundings. Heat exits the reaction. Surroundings warm up in temperature.

2H_2(g) + O_2(g) ----> 2H₂O_(l) + energy

H₂O_(g) ----> H₂O_(l) + energy

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) + O_2(g)

energy + H₂O_(s) ----> H₂O_(l)

energy can be converted from one form to another, but cannot be created or destroyed

ΔE = q + w

ΔE = q + w

ΔE = (-128J) + (+462J)

     = 334J

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

one which shows the enthalpy change as well as the mass relationships

N_2(g) + 3H_2(g) ----> 2NH_3(g)

ΔH = -91.8kJ

the sign of ΔH changes

CaCO_3(s) --Δ--> CaO_(s) + CO_2(g)    ΔH = +178kJ

CaO_(s) + CO_2(g) --Δ--> CaCO_3(s)    ΔH = -178kJ

2Ag₂S_(s) + 2H₂O_(l) --> 4Ag_(s) + 2H₂S_(g) + O_2(g)

ΔH = +595.5kJ

Given the equation above, calculate ΔH for the following reaction:

Ag_(s) + ¹/₂H₂S_(g) + ¹/₄O_2(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?

P_4(s) + 5O_2(g) --> P₄O_10(s)  ΔH = -3013kJ

(P₄ = 123.9g/mol)

P_4(s) + 5O_2(g) --> P₄O_10(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?

CH_4(g) + 1O_2(g) --> CO_2(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?

N_2(g) + 3H_2(g) --> 2NH_3(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

How much heat is given off when an 869g iron bar cools from 94^oC to 5^oC?

(Fe = 0.444J/g^oC)

q = C x m x ΔT

ΔT = 5^oC - 94^oC = -89^oC

q = (0.444J/g^oC) x (869g) x (-89^oC)

q = -34000J or -34kJ

How much energy must be transferred to raise the temperature of a cup of coffee (250mL) from 20.5^oC (293.7K) to 95.6^oC (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 99.8^oC

♥ Drop into 2225g water at 21.0^oC

♥ Water and metal come to 23.1^oC

♥ What is the specific heat capacity of the metal?

heat lost by metal = heat gained by H₂O

-q_metal = q_water

- (C) x m x ΔT = (C) x m x ΔT

- (C)(55g)(-76.7^oC) = (4.184J/g^oC)(225g)(21^oC)

C = .469J/g^oC

Suppose 0.562g of graphite is placed in a claorimeter with an excess of oxygen at 25.0^oC 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) + O_2(g) --> CO_2(g)

On reaction, the calorimeter temperature rises from 25.0^oC to 25.89^oC. 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 (25.89^oC - 25.0^oC)

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 25.0^oC, rises to 31.8^oC. 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/g^oC 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).

q_solution = C x m x ΔT

33mL + 42mL = 75mL x (1g/1mL) = 75g

q = (4.184J/g^oC) x (75g) x (31.8^oC - 25.0^oC) = 2133.84J

q_reaction = -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, HNO_3(aq) is ΔH^o_f = -207.4kJ/mol. Which equation is associated with this value?

a) H_(g) = N_(g) + O_3(g) --> HNO_3(aq)

b) ¹/₂H_2(g) + ¹/₂N_2(g) + ³/₂O_2(g) --> HNO_3(aq)

c) HNO_3(aq) --> ¹/₂H_2(g) + ¹/₂N_2(g) + ³/₂O_2(g)

d) HNO_3(aq) --> H_(g) + N_(g) + 3O_(g)

e) ¹/₂H_2(g) + ¹/₂N_2(g) + O_2(g) --> HNO_3(aq)

From the standard enthalpies of formation listed below, calculate the enthalpy change, ΔH^o, for the following reaction:

C₂H_4(g) + 3O_2(g) --> 2CO_2(g) + H₂O_(l)

Substance                      (ΔH^o_f)kJ/mol

C₂H_4(g)                           +52.3

CO_2(g)                            -393.5

H₂O_(l)                             -285.9

ΔH^o_rxn = ∑ΔH^o_f(products) - ΔH^o_f(reactants)

ΔH^o = [2mol (-393.5kJ/mol) + 2mol (-285.9kJ/mol)] - (52.3kJ/mol)

ΔH^o = -1411kJ

Calculate the standard form of enthalpy of formation of CS_2(l) given that:

C_(graphite) + O_2(g) --> CO_2(g)              ΔH^o_rxn = -393.5kJ

S_(rhombic) + O_2(g) --> SO_2(g)               ΔH^o_rxn = -296.1kJ

CS_2(l) + 3O_2(g) --> CO_2(g) + 2SO_2(g)    ΔH^o_rxn = -1072kJ

C_(graphite) + 2S_(rhombic) --> CS_2(l)

C_(graphite) + O_2(gas) --> CO_2(g)            ΔH = -393.5kJ

2S_(rhombic) + 2O_2(g) --> 2SO_2(g)          ΔH = 2x - 296.1kJ

CO_2(g) + 2SO_2(g) --> CS_2(l) + 3O_2(g)    ΔH = +1072kJ

C_(graphite) + 2S_(rhombic) -->CS_2(l)          ΔH = 86.3kJ