H2O is the Bronsted-Lowry acid in the forward reaction, donating an H+ to CH3, and CH4 is the Bronsted-Lowry acid in the reverse direction, donating an H+ to OH-.
Answer:
D. Up-Down
Explanation:
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<u>Answer:</u> The value of
for the reaction is 1051.93 J/K
<u>Explanation:</u>
Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.
The equation used to calculate entropy change is of a reaction is:
![\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28product%29%7D%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20S%5Eo_%7B%28reactant%29%7D%5D)
For the given chemical reaction:

The equation for the entropy change of the above reaction is:
![\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%5CDelta%20S%5Eo_%7B%28Cr_2O_3%28s%29%29%7D%29%5D-%5B%284%5Ctimes%20%5CDelta%20S%5Eo_%7B%28Cr%28s%29%29%7D%29%2B%283%5Ctimes%20%5CDelta%20S%5Eo_%7B%28O_2%28g%29%29%7D%29%5D)
We are given:

Putting values in above equation, we get:
![\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo_%7Brxn%7D%3D%5B%282%5Ctimes%20%28881.2%29%29%5D-%5B%284%5Ctimes%20%2823.77%29%29%2B%283%5Ctimes%20%28205.13%29%29%5D%5C%5C%5C%5C%5CDelta%20S%5Eo_%7Brxn%7D%3D1051.93J%2FK)
Hence, the value of
for the reaction is 1051.93 J/K
The IUPAC name of the compound <span>ch3–ch2–c ≡ c–ch3 is
PEN-2-YNE or
2-PENTYNE.
Attached below is a diagram that fully explains how the name was given and derived.</span>