25. Using the diagram below and the given data, calculate...

25. Using the diagram below and the given data, calculate ?H1, ?H2 and ?H4. H2 (g at 100°C) + 1/2 O2 (g at 100°C) ---------------------> H2O (liquid at 100°C) ?H3 1 ?H1 H2 (g at 25°C) + 1/2 O2 (g at 100°C) 2 ?H2 H2 (g at 25°C) + 1/2 O2 (g at 25°C) ---------------------> H2O (liquid at 25°C) ?H3 = -285.830 kJ/mol 3 Cp(H2) 25°C to 100°C = 28.91 J·deg?¹·mol?¹ Cp(O2) 25°C to 100°C = 29.41 J·deg?¹·mol?¹ Cp(H2O) 25°C to 100°C = 75.51 J·deg?¹·mol?¹ A. ?H1 = -2.168 kJ/mol ?H2 = -1.103 kJ/mol ?H4 = +5.662 kJ/mol B. ?H1 = -2.168 kJ/mol ?H2 = -1.103 kJ/mol ?H4 = -5.662 kJ/mol C. ?H1 = -2.168 kJ/mol ?H2 = -2.206 kJ/mol ?H4 = -5.662 kJ/mol D. ?H1 = +2.168 kJ/mol ?H2 = +1.103 kJ/mol ?H4 = -5.662 kJ/mol

Transcript

00:01 To figure out delta h1 h2 and delta h4 so the first reaction delta h1 is and you can see the change is a hydrogen molecule in a gas form at 100 degrees celsius turn into a hydrogen ion still gas but now it's 25 degrees celsius.

00:30 Now according to the bottom information, hydrogen, when it increased from 25 to 100, the calorie absorbed is 28 .9 9, dual per degree celsius per mole.

01:02 Now in our case, you can see it's the opposite around.

01:05 So you have 100 going down to 25.

01:09 So the information given was 25 to 100.

01:13 So obviously in our case, this is the opposite process.

01:17 Instead of absorb 28 .9, it released 28 .9.

01:23 So from there, you can see that the delta delta h1 is going to be a negative value it's a release heat negative and then you can see that it's 28 .9 dual per degree celsius per mole and in our case you go from a hundred degree celsius to 25 so the delta t the change of degree is 75 so from there you can see that delta h1 is going to be negative 28 .9 joule per degree celsius per mole multiplied by 75 degrees celsius and this will give you negative 2 ,168 and it's a negative negative there again this is because you can see that we have 25 to 100 which absorb heat and this is the opposite process it's going to be released for process so negative now let's move on to the delta h2 so in this case you have half more of oxygen that is a guess at a hundred degrees celsius and it become a half more of oxygen still at 25 degrees celsius again according to the information that we were given the heat oxygen absorbed from 25 degree to a hundred degree is actually 29 .4 joules per degree per mole.

03:13 So you can see that according to what we have for each mole, when you go from 25 to 100, it absorbed 29 .4 joules per mole per degree.

03:36 But in our case, we have half.

03:41 So this means that the delta h2, again, you go from the opposite direction, 100 back to 25 instead 25 to 100.

03:52 So again, you're gonna have to put a negative sign here because it's the reverse reaction, reverse process when you have 100 to 25, which means the heat will be released.

04:04 So negative 29 .4, u .b per degree per mole, multiply by half, because you have half, more and again the temperature change is 75 so this tells us that the delta h2 is negative um 1 .103 kjou per mole now as you can see i'm just being lazy because it's negative 1103 k jr or sorry joel per mole um joel is uh 1 000 but you're right down as a k -jol, kilojole.

05:00 So basically, this number divided by a salary of 1 .103 kajou per more.

05:06 Now, let's move on to the delta h4...

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