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;
Iodine
Explanation;
Electron configurations are a way of keeping track of the location of the electrons around the nucleus.
Iodine is an element which belongs to the halogen family. The halogen group includes fluorine, chlorine, bromine, astatine, and iodine.
A neutral iodine atom would also have 53 electrons. Its ground state electron configuration would be:
1s22s22p63s23p64s23d104p65s24d105p5
Answer:
2l- ---> l2 + 2e- is the anode
2H+ + 2e- ---> H2(g) is the cathode
Explanation:
Oxidation occurs when a metal loses two or more electrons in a redox chemical reaction and reduction is when it gains. Thus, oxidation is the anode and reduction is the cathode.
