3.1 Physical Chemistry: .4 Energetics Flashcards Preview

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Flashcards in 3.1 Physical Chemistry: .4 Energetics Deck (28):
1

what is an exothermic reaction?

a reaction where heat energy is given out to the surroundings

to make bonds

there is usally a temperature increase

2

what is an endothermic reaction?

a reaction where heat energy is absorbed (taken in) from the surroundings

to break bonds

the temperature usually decreases

3

in an exothermic reaction is ΔH positive or negative?

negative

4

in an endothermic reaction is ΔH positive or negative?

positive

5

In a endothermic reaction out of the reactants and products which one has more energy?

the products have more energy than the reactants

6

In a exothermic reaction out of the reactants and products which one has more energy?

the reactants have more energy than the products

7

what is meant by enthalpy change? what are the units

enthalpy change is the heat energy transferred in a reaction at consant pressure

units: kJ mol-1

8

what are the standard conditions for measuring the enthalpy changes of a reaction?

-Pressure: 100kPa

-Temperature: 298 K (25 degrees)

-1 mole

9

why is it important to have a standard of pressure for measuring energy changes?

-pressure affects the amount of heat energy given out by the reactions that involve gases

-if a gas is given out. some energy is required to push away the atomsphere

-the greater the atomspheric pressure the more energy is used for this

-this means that less energy remains to be given out as heat by the reaction. `

10

what is the eqaution for calculating ΔH (the enthalpy or heat energy change)?

 

ΔH= Σ Er (sum of bonds broken) - Σ Ep (sum of bonds made)

Σ= sigma (sum)

Er= Energy of reactants

Ep= Energy of products

11

Calculate the overall enthalpy change for this reaction:

N2(g) + 3H2(g) → 2NH3(g)

Use the mean bond enthalpy valules

N=N 945 kJ mol-1

H-H 436 kJ mol-1

N-H 391 kJ mol-1

Q image thumb

+ΔH  (bonds broken)

(N=N x 1)= 945 kJ mol-1

(H-H x 3)= 1308 kJ mol-1

Total= 2253 kJ mol-1

 

-ΔH (bonds made)

(N-H 391 x 6)= 2346 kJ mol-1

Total= -2346 kJ mol-1

 

ΔH= Σ Er - Σ Ep

ΔH= 2253 - 2346

ΔH= -93 kJ mol-1

 

 

12

what is bond dissociation energy?

the enthalpy (heat energy) change when 1 mole of a particular covalent bond is broken, all species are in the gaseous state

under standard conditions of;

298 K

100kPa

1 mole

13

give the equation for the bond dissociation of chlorine

Cl2(g) → 2Cl(g)

14

what is the definition for the enthalpy of formation? ΔHfθ

this is the enthalpy (heat energy) change when 1 mole of a compound is formed from its elements, all reactants and products in their standard states when measured under standard conditions of 298 K and 100 kpa

15

give the eqaution where hydrogen, nitrogen and oxygen are reacted together to make 1 mole of HNO3 (in enthalpy of formation)

1/2H2 + 1/2N2 + 3/2O2 → HNO3

16

what is the definition for the enthalpy of combustion?ΔHcθ

This is the enthalpy (heat energy) change when 1 mole of a substance is burnt in excess oxygen, all reactants and products in their standard states when measured under standard conditions of 298 K and 100 kPa

17

for the eqaution:

4Na(s) +  O2(g) → 2Na2O(s)  

state the formula

you would use for the enthalpy of combustion

ΔHθ= 4 x (ΔHcθ [Na(s)]

18

state the eqaution you would use to work out the heat change in a reaction

q= mcΔT

m= mass/grams  (usually gives you in cm3 which = g)

c= specific heat capacity (for water its usually 4.18)

ΔT= temeperature change in Kelvin

19

give an eqaution for the buring of Sodium in excess oxygen to make sodium oxide (enthalpy of combustion)

1Na(s) +  1/4O2(g) → 1/2Na2O(s)  ΔHcθ (Na)

(you are only allowed to make 1 mole of Na because of the standard conditions)

(you get 1/4 O2 because 1/2 O produced divided by 2 O used gives a 1/4)

20

describe a experiment where you use calorimetry to find out how much heat is given out by a reaction

-state the equipment needed

equipment needed:

-stirrer

-thermometer

-Fuel (reactant)

-water

-copper calorimeter

1. weight the intial mass of your feul and note it down

2. measure the temperature of the water before starting to burn and note it down

3. heat the water to a temperature change of 20 degrees (or ΔT of your choice)

4. measure the final mass of feul

 

21

In a labortory experiment, 1.6 g od an organic liquid fuel were completely burned in oxygen. The heat formed during this combustion raised the temperature of 100g of water from 295 K to 358 K. Calculate the standard enthalpy of combustion ΔHcθ, of the fuel. Its Mr is 58.0

 

Q image thumb

q=mcΔT

q= 100 x 4.18 x (358-295)= 26 334 J

convert jules into kilojoules (26 334 J/ 1000)= 26.334 kJ

 

then find out how many moles of fuel produced

number of moles of fuel = 1.16/58.0 = 0.0200

so the heat produced by 1 moles of fuel (ΔHcθ)= -26.334/ 0.0200

= -1320 kJ

you put a minus infront of the 26.334 because the reaction is exothermic. you know this since it raised the temperature of the water

22

what are the experimental problems with all calorimetry?

-some heat will be absorbed by the container, rather than going towards heating up the water

-some heat is always lost to the surroundings during the experiment (however well the container's insulated) 

23

what are the experimental problems with flammable liquid calorimetry?

-some combustion may be imcomplete- which will mean less energy will be given out

-some of the flammable liquid may escape by the vaporation (they're usually quite volatile)

24

give the definition for Hess's Law

the enthalpy chnage at constant pressure is independan of the route.

The intial and final energu states are constant

25

work out the enthalpy change of combustion to calculate ΔHrθ of ethanol (C2H5OH)

using the thermochemical cycle

ΔHcθ [C(s)]= -394 kJ mol-1

ΔHcθ [H2(g)]= -286 kJ mol-1

ΔHcθ [C2H5OH]= -1367 kJ mol-1

start with your equation:

2C(s) + 3H2(g) + 1/2O2(g) --> C2H5OH(l)

use the equation  ΔH1=  ΔH2 +  ΔH3

 

A image thumb
26

what is the mean bond enthalpy?

this is the average or mean change when 1 moles of a particular covalent bond is broken. when averaged over many different environments, all speices are gaseous state.

27

explain why values from mean bond enthalpy calculations differ from those determined using Hess's law

-the values determined by Hess's law is the 'correct value' because all the enthalpy of formation changes of the values have been obtained from the actual compunds involved

-mean bond enthalpy calculations also allow us to calculate an approximate value for ΔHfθ for a compound that has never been made.

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