00:01
This question is asking us to calculate the free energy of this equation.
00:08
It has provided us these equations and their respective free energy values.
00:14
To start, let's mark these equations as one, two, and let's look at what we have.
00:32
So the question asks for carbon, equation 2 has carbon, hydrogen, equation 3 has hydrogen, and c6h6, equation 1 has c686.
00:46
So let's begin with carbon.
00:49
We need six carbon atoms.
00:52
So let's multiply equation 2 by 6.
01:04
We get 6 molds of carbon, 6 of oxygen, 6 of co2.
01:11
And we also need to multiply the free energy by 6.
01:14
So 394 times 6 will give us minus decrease, six four kilojoules.
01:23
Next is hydrogen.
01:25
We need three hydrogen.
01:34
So let's take this equation, equation three, and multiply it by three.
01:50
Free energy for this equation would now be 7, 11 kilojoules.
01:58
And let's multiply this by three, another three, and three.
02:08
Okay, now let's add these two.
02:10
Equations.
02:12
So that would be 6c plus 3h2 plus 6 o2, 6 plus 3 by 2 is 1 .5, so 6 plus 1 .5, 7 .5 .02, and 6c02 and 3h2...