This question is describing the following chemical reaction at equilibrium:
And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:
Thus, by recalling the Van't Hoff's equation, we can write:
Hence, we solve for the enthalpy change as follows:
Finally, we plug in the numbers to obtain:
![\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Cfrac%7B-8.314%5Cfrac%7BJ%7D%7Bmol%2AK%7D%20%2Aln%280.25%2F9%29%7D%7B%5B%5Cfrac%7B1%7D%7B%2875%2B273.15%29K%7D%20-%5Cfrac%7B1%7D%7B%2825%2B273.15%29K%7D%20%5D%20%7D%20%5C%5C%5C%5C%5C%5C%5CDelta%20H%3D4%2C785.1%5Cfrac%7BJ%7D%7Bmol%7D)
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Answer:
4KNO3 ==> 2K2O + 2N2 + 5O2
Explanation:
It's a decomposition, but not a simple one.
KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3
4KNO3 ==> 2K2O + 2N2 + 5O2
Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2
That should do it.
I think it is because it is directly about the north or south pole (I forget) so while all the other stars seem to rotate around because the Earth is rotating but the North Star is directly in the center. Don't take my word though.
