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Entry Task: May 22 nd 23 rd Block 1. Question: What is temperature change of 50.0 grams of water absorbs 35kJ of energy. (specific heat 4.18 J/ gC ) You have 5 minutes!. Agenda:. Discuss Ch. 16.sec 2 and worksheet In-class (small notes) on Sec. 3 HW:

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Entry Task: May 22nd 23rd Block 1

Question:

What is temperature change of 50.0 grams of water absorbs 35kJ of energy. (specific heat 4.18 J/gC)

You have 5 minutes!

Agenda:

Discuss Ch. 16.sec 2 and worksheet

In-class (small notes) on Sec. 3

HW:

Look at the specific heats of metal Lab

A calorimeter is an insulated device used for measuring the amount to heat absorbed or released during a chemical or physical process.

Define a calorimeter

A known mass of water is placed in an insulated chamber.

• The chamber contains the reaction- any heat absorbed in the reaction will drop the temperature of the water.
• Any heat released in the reaction will raise the temperature of the water.
How does it work?

q= cmΔT

• 9750 J = (4.184 J/g˚C)(335g)(X)
• 9750

(4.184)(335)

1401.64

6.96 ˚C is the temperature difference

65.5 - 6.96 = 58.5 ˚C

If 335g of water at 65.5˚C loses 9750 J of heat, what is the final temperature of the water?

q= cmΔT

• 5650 J = (4.184 J/g˚C)(Xg)(26.6˚C)
• 5650

(4.184)(26.6)

111.29

= 50.8 g

The temperature of a sample of water increases from 20.0˚C to 46.6˚C as it absorbs 5650 J of heat. What is the mass of the sample?

Chaos

Most energy

Taking in energy

(endothermic)

Releases energy

(exothermic)

Taking in energy

(endothermic)

Releases energy

(exothermic)

Taking in energy

(endothermic)

Remember this?

Order

Less energy

Less order

More energy

Releases energy

(exothermic)

The system is the reaction

• The surroundings is the everything other than the reaction.
• The universe is the system AND surroundings.

Surroundings

UNIVERSE

System

• *its almost impossible to know the absolute amount of energy held in a substance.
• Enthalpy (heat) of reaction is the change of energy- one substance to another.
• ΔHrxn = Hfinal – Hinitial
What is the difference between enthalpy and the enthalpy (heat) of reaction?

The sign of the enthalpy reactions

For an exothermic reaction, heat energy is _____________.

When the Hreactants is subtracted from the ________Hproduct, a ________ value for ΔHrxn is obtained.

Enthalpy changes for exothermic reactions are always_______________.

released

smaller

negative

negative

4Fe (s)

+

O2 (g)

2Fe2O3 (s)

ΔH= -1625 kJ

Notice the phase change

(s) + (g)  (s)

Gas to solid loses energy- exothermic

Disorder to order = lose energy

The sign of the enthalpy reactions

For an endothermic reaction, heat energy is _____________.

When the Hreactants is subtracted from the ________Hproduct, a ________ value for ΔHrxn is obtained.

Enthalpy changes for endothermic reactions are always_______________.

absorbed

larger

positive

positive

NH4NO3(s)

NH4+ (aq)

+

NO3- (aq)

ΔH= 27 kJ

Notice the phase change

(s)  (aq) + (aq)

Solid to liquid gains energy- endothermic

Order to disorder gains energy

Identify the reaction as endothermic or exothermic-PAY ATTENTION TO PHASE CHANGES

Endothermic

C2H5OH (l)  C2H5OH(g)

NH3(g)  NH3(l)

Br2(l)  Br2(s)

NaCl (s)  NaCl (l)

C5H12 (g) + 8O2 (g)  5CO2 (g) + 6H2O(l)

Exothermic

Exothermic

Endothermic

Exothermic

B

C

A

F

D

E

G

9. What are the surroundings?

10. What is the universe?

The solution

Solution + surroundings

Thermochemical equations

Are balanced chemical equations that include physical states of all reactants and products AND the energy change ∆H.

NH4NO3(s)

NH4+ (aq)

+

NO3- (aq)

ΔH= 27 kJ

Types of enthalpy

∆H(comb)

∆H(vap)

∆H(fus)

Change in heat with combustion

Change in heat with vaporization

Change in heat with fusion

Thermochemical equations for changes in state

Consider:

H2O(l)  H2O(g)

ΔH(vap)= 40.7 kJ

What is happening?

Its vaporizing.

What ∆H would you use? Would it be + or -?

Positive- endothermic

Thermochemical equations for changes in state

Consider:

H2O(s)  H2O(l)

ΔH(fus)= 6.01 kJ

What is happening?

Its melting.

What ∆H would you use? Would it be + or -?

Positive- endothermic

Pay attention to the signs

∆Hvap= -∆Hcond

∆Hfus= -∆Hsolid

So -∆Hcond for water is

H2O(g)  H2O(l)

ΔH(cond)= - 40.7 kJ

So -∆Hsolid for water is

H2O(l)  H2O(s)

ΔH(fus)= - 6.01 kJ

Makes sense- both are exothermic -∆H

Calculating energy released in a reaction

How much heat is released when 54.0 g of glucose (C6H12O6) is burned?

C6H12O6 (s) + 6O2(g) 6CO2(g) + 6H2O(l)∆H comb= -2808 kJ

∆H comb= -2808 kJ is per mole of glucose

Convert gram to mole

1 mole glucose

54.0g glucose

= 0.3 mol

180 g glucose

= 0.3 mol X -2808 ∆H comb =842 kJ

You Try!!

1. Calculate the heat required to melt 25.7 g of solid methanol at its melting point.

Which ∆H would you use?

∆H (fus) because its melting. So 3.22 its is.

∆H(fus) be positive or negative?

∆H (fus) be positive. ∆H (fus)= 3.22 per mole.

You Try!!

1. Calculate the heat required to melt 25.7 g of solid methanol at its melting point.

∆Hfus= 3.22 kJ is per mole of methanol

Convert gram to mole

1 mole methanol

25.7g methanol

= 0.803 mol

32 g methanol

= 0.803 mol X 3.22 ∆H fus=2.58 kJ

You Try!!

2. How much heat is evolved when 275g of ammonia gas condenses to a liquid at its boiling point?

Which ∆H would you use?

∆H (vap) because its condensing. So 23.3 its is.

∆H(vap) be positive or negative?

∆H (vap) be negative. ∆H (vap)= -23.3 per mole.

You Try!!

2. How much heat is evolved when 275g of ammonia gas condenses to a liquid at its boiling point?

∆Hvap= -23.3 kJ is per mole of ammonia

Convert gram to mole

1 mole ammonia

275g ammonia

= 16.2 mol

17 g ammonia

= 16.2 mol X -23.3 ∆H vap=377 kJ

You Try!!

3. What mass of methane must be burned in order to liberate 12880kJ of heat? Methane (CH4) ∆Hcomb= -891 kJ/mol

12880 ÷ -891 =14.5 moles of methane

16 g methane

14.5molmethane

= 232 g

1 mole methane

You Try!!

2. How much heat is evolved when 275g of ammonia gas condenses to a liquid at its boiling point?

∆Hvap= -23.3 kJ is per mole of ammonia

Convert gram to mole

1 mole ammonia

275g ammonia

= 16.2 mol

17 g ammonia

= 16.2 mol X -23.3 ∆H vap=377 kJ