Ch4g 2o2g co2g 2h2oliq 213 kcal using the bond data in table...

CH4(g) + 2O2(g) ===> CO2(g) + 2H2O(liq) + 213 kcal Using the bond data in Table 7.4.1 in the course textbook (Link to section containing table 7.4.1Links to an external site.) show that the stated heat produced by the reaction is consistent with change in enthalpy between the reactants and products for the reaction. Hint: You can assume the following for the number and types of bonds broken and formed: Bond Type Broken Formed C-H 4 O=O 2 C=O 2 O-H 4

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00:01 We are asked to consider the following chemical reaction.

00:03 I'm going to go to black and 2h2o.

00:21 That says it's a liquid and we're told that our heat is negative 213 kilocalories.

00:33 We are asked to use the bond data in the text, which i don't have.

00:47 So you're going to have to verify the bond energies.

01:12 I use match years.

01:30 There's something very nice here.

01:32 We are given bonds broken and bonds formed.

01:51 So we're given this information.

01:56 So our bonds broken will be 4ch bonds and 2, i'm going to get rid of this.

02:12 It's a little confusing.

02:17 O double bond o.

02:21 And we are forming 2c double bond o and 4oh.

02:33 Again, i don't have the same as yours.

02:38 I wonder if i have kilocalories.

02:39 I do.

02:41 So the numbers that i'm going to try to match, i'm going to do in red.

02:46 Ch is 100.

02:48 That'll be 400 kilocalories.

03:07 And my o double bond o is 19.

03:11 And that will be 200 kilocalories, 238 kilocalories plus 400 kilocalories equals 638 kilocalories minus here, where's my c double bond o? there we go.

03:52 190 is 380 and oh is 110.

04:18 So 380 plus 440 equals 820.

04:32 Oopsie, what am i doing here? i got an ugly thing in my calculator...

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