Answer:-ΔG=-101.5KJ
Explanation:We have to calculate ΔG for the reaction so using the formula given in the equation we can calculate the \Delta G for the reaction.
We need to convert the unit ofΔS in terms of KJ/Kelvin as its value is given in terms of J/Kelvin
Also we need to convert the temperature in Kelvin as it is given in degree celsius.
After calculating forΔG we found that the value ofΔG is negative and its value is -101.74KJ
For a reaction to be spontaneous the value of \Delta G \ must be negative .
As the ΔG for the given reaction is is negative so the reaction will be spontaneous in nature.
In this reaction since the entropy of reaction is positive and hence when we increase the temperature term then the overall term TΔS would become more positive and hence the value of ΔG would be less negative .
Hence the value of ΔG would become more positive with the increase in temperature.
So we found the value of ΔG to be -101.74KJ
x= the coefficients in front of the substance in the balanced chemical equation
[H+]= the concentration of hydrogen ions
[A-]= the concentration of the other ion that broke off from the H+
[HA]= the un-disassociated acid concentration
The higher the Ka value, the greater amount of disassociation of the reactants into products. As for acids, they will break down to form H+ ions. The more the H+ ions, the stronger acidity of the solution. Thus since A has the highest Ka value, that represents the strongest acid.
You can determine the Ka value from a number of ways. If equilibrium concentrations are given of a certain acid solution, you can find the proportion of the concentration of ions to the concentration of the remaining HA molecules, using the equation above. Also, pH and KpH can be used in a number of ways. This gets more complicated and depends on the situation, and requires more advanced equations.
Hope this helped a little, its obviously not my best work
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Answer:
0.4 M
Explanation:
The process that takes place in an aqueous K₂HPO₄ solution is:
First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:
- 200 mL * 0.2 M = 40 mmol K₂HPO₄
Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:
- 40 mmol K₂HPO₄ *
= 80 mmol K⁺
Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:
- 80 mmol K⁺ / 200 mL = 0.4 M
