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Calculate Hrxn for the reaction:

Use the following reactions and given H values:

$$-173.2 \mathrm{kJ}$$

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

Rice University

University of Kentucky

Brown University

Lectures

00:42

In thermodynamics, the zer…

01:47

A spontaneous process is o…

03:48

Calculate Hrxn for the re…

02:50

03:23

07:15

Calculate Grxn for the r…

00:52

Use standard enthalpies of…

06:50

00:47

Calculate $\Delta H_{\math…

03:37

01:58

00:44

01:21

Consider the generic react…

03:00

Rewrite each equation belo…

00:56

Write the equation for cal…

02:41

Determine G for the reac…

02:08

0:00

Use bond enthalpies in Tab…

04:17

Identify the reagents (a$-…

Calculate $\Delta H^{\circ…

Use bond energy values (Ta…

01:19

Calculate $K_{e q}$ for th…

So in this question we have this reaction x, 5 common plus 6 hydrogen to give 5 to liquid, and we are told to calculate the delta h of reaction for this particular equation. But we have given some 100 equations. Like c f, i t plus oxygen 5 c, o 2 plus c squatter and the other 1 of tastes. So we are set to use these 3 equations to provide an answer to the delta h of the reaction of this first question. So all we go about doing this, all we need to do is to check the equation of reaction and made show that the, for example, the 5 kalon is on the reactant side and the simple hydrogen and the 2 is on the product side. So if we check that the 5 colon we have common here, we have common on the reactant side. In the second equation. We have hydrogen on the reactants. We have hydrogen on the reactant starting with third equation and potch 2 is on the product side, which is on the reactant side in this first equation. So that means we're going to need to sleep this first equation and then looking at these, we have a 5 column where here we have just 1. So we need to multiply this second equation like 5 and we need just at to apply the that equation by 3, because we have 200 here- and we have 6 years so multiplied by 3 to make 6 we're just going to go to the next page and Do that to remember, we need to leave this first equation, say: eli, we'll have a 5 c plus c h, 2 thought plus oxygen, and whenever you leave this sign for the delta of the reaction will change it well. So previously it was minus 3 to 44.8. So right now it's going to be plus 321.8 kilo joules. So the second equation. We need to multiply by 5. So just go ahead and you get so that will be 5 c plus 5 o 2 to 52 and we got to. But the reaction is minus s 167, in 67.5 kilo joules and for the last equation. We need to act, apply by 3 c, h, 2 plus 3 for 2 to give c sah 2 and then the delta h for the reaction will be. We multiply the original value by 6 as by 3, so that is going to be 43.5 times 3, so that is minus 1450.5 kilo joules. So if we sum everything up, 5 c, o 2 cancel 5 c, o 2 ch 2 cancel c h, 2, o 5 plus 3. That'S equal to or cancel as the will be lengthis, 5 c, plus c 2. To give c 512, which is the same tenacity situation, i was started with ay that we want to find so that is delta. H, then, will be equals to 324418 minus 1967.5 minus 1.5, 0.5 and i'll be minus 173.2 kilo. Joules! That'S the answer for the question.

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