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00:00:05

hello guys, my name is

00:00:07

Tatyana Belousova, a brilliant chemistry teacher, today you

00:00:10

and I will continue to learn how to solve

00:00:13

problems using Hess’s law, and today we

00:00:17

will learn how to make calculations when there is a so-

00:00:20

called direct path to obtaining

00:00:23

certain substances, but when such a

00:00:27

reaction is not possible directly, but we need

00:00:29

solve the calculation, do it along an indirect path of

00:00:34

transition, so the first task will

00:00:37

look like this:

00:00:39

we are asked to calculate what the

00:00:43

thermal effect of the reaction will be

00:00:45

if we have the formation of

00:00:51

nitrogen oxide 5 n 2 o 5 I’ll immediately make a reservation that

00:00:56

such a reaction is directly impossible nitrogen

00:00:59

when it reacts with oxygen it forms

00:01:02

everything - after all, nitrogen monoxide also wants

00:01:05

difficult conditions under the influence of lightning

00:01:07

discharges or, for example, high

00:01:10

temperature, which means we are offered the

00:01:12

following chemical transformation paths

00:01:17

that we can use two and but

00:01:20

this is gas plus two is gas and the result

00:01:26

is 2 and by 2 this is gas and

00:01:31

delta h of this transition is minus

00:01:36

114 barely kilojoules then there is 2 transition

00:01:43

4

00:01:44

no2 is a gas plus a-two

00:01:48

is a gas and it turns out two n 2 o 5 n 2 o 5

00:01:55

is a solid substance it turns out

00:01:58

this transition delta h 2 has -110

00:02:03

and 2 kilojoules

00:02:08

and there is a transition 3 n 2 is a gas plus a-two is

00:02:15

also a gas and it turns out 2 and but

00:02:19

being a gas, delta h 3

00:02:23

is 182 and 6 kilojoules and so

00:02:29

using these three chemical

00:02:31

equations we need to find a way to

00:02:35

calculate so that we get the heat of

00:02:38

such a transition n2 + 5 2 x 2 it turns out

00:02:43

n 2 o 5 let's see how this is

00:02:45

done and so firstly in these

00:02:48

equations you can notice such

00:02:50

substances that we do not have in our

00:02:53

equation

00:02:54

include them and but also on 2 for example

00:02:57

which are not in this interaction then

00:03:00

we do the following So, let's take the

00:03:05

first two equations, let's see in

00:03:09

these equations there is the same

00:03:11

substance no2,

00:03:12

which should ultimately be reduced, and I

00:03:15

draw your attention to the right side of us,

00:03:18

eat exactly the same substance n 2

00:03:21

o 5, which we need for

00:03:24

interaction, but between the first

00:03:27

equation where there are two and on 2 and the second

00:03:30

where there is 4 on 2 there is an inconsistency in this regard;

00:03:32

moreover, before n 2 o

00:03:36

5 in our equation there should be

00:03:38

units then look what we will do, we will

00:03:41

divide this equation by 2,

00:03:44

I mean, accordingly, this

00:03:47

equation in terms of coefficients for,

00:03:50

respectively, the thermal effect will also

00:03:52

change,

00:03:53

which means the first equation remains the same, I will

00:03:56

rewrite it as

00:03:57

2 and but I

00:03:59

will not write down the state of aggregation now, I will leave it as

00:04:03

is so

00:04:04

that we can do the calculation faster and,

00:04:07

accordingly, I will reduce the second equation to the

00:04:10

amount of 2 and by 2 plus a- two

00:04:15

equals respectively n 2 o 5

00:04:20

but in this case our delta h 2

00:04:23

will change and it will be

00:04:26

divided by two and will be minus 55 and 1

00:04:32

kilo jul

00:04:34

so we divided the second equation by 2

00:04:37

we get 2 and by 2

00:04:40

gas + 1 2 1 we divided for two gas

00:04:46

is equal to n 2 o 5 solid and the third equation

00:04:53

remains unchanged n2 gas plus a-two

00:04:57

gas it turns out 2 and well now look

00:05:01

what we are doing we add up all the left

00:05:04

sides and all the right sides to check

00:05:07

it will turn out for our equation as

00:05:09

given

00:05:10

so two and to gas plus a-two gas plus from the

00:05:17

second equation 2 to 2 gas + 1 2 o 2

00:05:25

gas plus from the third equation n2

00:05:29

gas plus a-two

00:05:32

gas from the right side there are two left no2

00:05:38

gas + n 2 o 5 solid and plus 2 and Well,

00:05:45

in the form of gas, now we’re trying to reduce it, it will

00:05:49

take us two by two by two by 22 and by

00:05:54

2 and what’s left is let’s see what we

00:05:58

have left in the end is n2

00:06:01

I’ll write it forward gas now with

00:06:04

oxygen carefully about 21 another

00:06:07

half 1 2 and another o2 will just be

00:06:11

a plus accordingly, 5 second o 2 well, and on the

00:06:17

right side of the country, as a result of the reduction, we

00:06:19

got

00:06:20

just n 2 o 5 which is

00:06:24

hard

00:06:25

and as we see our equation turned out

00:06:27

exactly as it was

00:06:30

originally given, but we still have to calculate our

00:06:33

numbers and so we’ll do it this way

00:06:36

delta h reactions be added up

00:06:40

delta h 1 plus delta h 2, taking into account the fact

00:06:45

that we divided by 2 and plus delta h 3

00:06:50

and make the substitution 1 minus then 14 barely

00:06:56

kilojoules so plus but we have a

00:07:01

minus sign here and we also divided minus

00:07:04

55 and 1 kilo jul

00:07:07

and accordingly in the last equation

00:07:11

plus 180 2 and 6 kilojoules

00:07:16

if we calculate all this figure all these

00:07:19

numbers we get respectively

00:07:22

plus 13 and 3 kilojoules

00:07:27

means in this case again n2 is a gas

00:07:32

o2 is a gas n 2 o 5 solid

00:07:35

we got a delta h reaction plus

00:07:41

13 and 3 kilojoules, respectively, namul,

00:07:46

but then q will have a minus sign

00:07:50

and, accordingly, it will be minus 13 three kilo

00:07:55

joules icons, which means this process

00:07:58

will be indo

00:08:00

thermally, I remind you once again that delta

00:08:03

h we have kilo john per mole and

00:08:06

accordingly we have solved the first problem and

00:08:09

proceed to the second problem we will continue

00:08:12

solves we need to calculate

00:08:15

the heat of formation board 26 this

00:08:20

substance is called deboron

00:08:22

from boron and hydrogen I want to immediately

00:08:25

clarify that such a reaction directly is

00:08:27

practically impossible for this we are

00:08:30

offered three stages of converting

00:08:33

boron and hydrogen into board 26

00:08:37

so we start with the same thing with analysis

00:08:39

let's look at bar 2 a6 in the first

00:08:43

equation of the reaction is on the left and in

00:08:46

our final equation it should be

00:08:49

on the right so the first thing we do is

00:08:51

transfer this equation and swap

00:08:54

places from left to right, which allows us to

00:08:57

do a consequence of the law of weight under number

00:09:00

3 port 2 dry solid plus 3 h2o which

00:09:08

in this case is a gas goes

00:09:12

into bar 26 gas and 3 a 2 gas but look

00:09:20

since we changed the

00:09:22

left side and the right side in this case, our delta h here

00:09:26

will change

00:09:27

delta h 1 will become with a plus sign 2035 and

00:09:33

6 kilo jul

00:09:35

then the second equation the second equation

00:09:39

we leave as there are two boron

00:09:42

solid plus 3y second o 2cos we get

00:09:48

boron 23

00:09:50

solid means in the last level now

00:09:55

let's see here water was formed

00:09:57

where there is water in our equations

00:10:00

water is in the first equation

00:10:02

but here and and 3 mol and in our last

00:10:05

equation it is

00:10:07

this substance with the amount of substance one,

00:10:10

then we need to multiply the last equation

00:10:13

by 3, respectively, we

00:10:16

get 3 h2 gas plus three second

00:10:23

o2 gas and, accordingly, 3 h2o will also be in the

00:10:29

form of gas on the right, but since we multiplied by

00:10:32

3, the correspondence

00:10:34

to in our case, delta h 3 will also

00:10:37

be multiplied by 3 and will accordingly be

00:10:40

equal to minus 240 1 and 8 multiplied by three

00:10:45

kilojoles, which we will take into account in the subsequent

00:10:49

calculation, now we add in the same way as we

00:10:52

did in the first problem, the left side

00:10:54

separately, the right parts separately observe the

00:10:57

same naturally the coefficients are

00:10:59

2 o 3 solid plus 3

00:11:04

h2o in the form of gas from the first reaction we took the

00:11:09

left side

00:11:10

+ 2 solid boron plus three second

00:11:15

o2 gas from 3 reactions 3

00:11:21

h2 gas plus three second

00:11:26

o2 gas the left side is all over,

00:11:30

let's move the right side here

00:11:33

and see what we have left on the right is

00:11:35

boron 26

00:11:37

gas plus 32

00:11:41

gas plus boron

00:11:44

2 o 3 solid plus 3 h2o in the form of gas

00:11:51

we give a similar one and so bar 2 o 3 and bar

00:11:56

23 we reduce 3

00:11:58

h2o and 3 h2o we reduce now with

00:12:03

oxygen let's look on the right

00:12:05

we have 3 a 2 here three second and three second

00:12:08

obviously one and a half and one and a half is also 3

00:12:11

then this gives us the right to reduce

00:12:13

oxygen too and we get the equation 2

00:12:18

boron solid plus 3 h2 gas we get boron

00:12:27

26

00:12:29

gas and it turns out our equation

00:12:32

corresponds to what we needed to

00:12:35

get that is ours here is the type of

00:12:37

movement of the equation of the reaction of ladies to

00:12:39

wives are correct, it remains to

00:12:41

calculate the delta h reaction

00:12:44

for this we use the legal gis and

00:12:47

say that we need to add delta h 1

00:12:50

and pay attention when we

00:12:52

turned this number over because we

00:12:54

had to do this in the equation plus

00:12:57

delta h 2 plus delta h 3,

00:13:03

respectively, multiplied by 3 since

00:13:06

we multiplied here delta h 1

00:13:09

so we don’t forget, we take the inverted thousand

00:13:11

inverse value of this value,

00:13:14

substitute it will be plus

00:13:18

2035 and 6 kilojoules now the second delta

00:13:25

h pay attention to the minus sign it

00:13:27

means minus 1273

00:13:31

and 5 kilos jor and now the third

00:13:35

digit remains -240 18 when we multiply by 3 it

00:13:40

will be minus 725 and 4 kilos jul and when we

00:13:47

do all the calculations you will

00:13:50

end up with a figure of 36.7

00:13:56

kilo jolt respectively all of this by 10 in

00:14:01

the calculations, the acu in our reaction will have the

00:14:05

opposite sign minus 3 6 and 7 kilo

00:14:09

jul

00:14:11

now this gives us the right to

00:14:13

add minus thirty six

00:14:17

and seven kilo jul to this equation, which means this reaction

00:14:21

will be n to thermal, this is how

00:14:28

problems are solved according to Hess’s law and as you

00:14:31

see, today in these problems we

00:14:34

multiplied the equation accordingly by a

00:14:36

certain factor and swapped the

00:14:39

left and right sides and, accordingly,

00:14:42

Hess's law thus allows us to

00:14:44

do such calculations, I hope that it

00:14:48

was clear to you and you will be able to solve a similar problem in the future

00:14:51

Kisame, goodbye guys

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