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I tried doing these questions and just wanted to know if there are any that I should look back at and fix. Please don't edit this. This looks exactly how it looks on my paper and in the book I am using.

The link for the part that says for example "See equation 11.1" and "See equation 11.5" is:

https://books.google.com/books?id=X3KX6p8hnlIC&pg=PA304&lpg=PA304&dq=Five+repeated+calculations+are+required+to+establish+the+data+plot+for+the+determination+of&source=bl&ots=AenLh4IGsI&sig=hs7pdP6AnhvgaTVs0zvcGChbTzE&hl=en&sa=X&ved=0ahUKEwjzpZ_WlKrJAhWBHD4KHQ-ICqIQ6AEIHDAA#v=onepage&q=Preparation%20of%20Borax%20solutions&f=false

I just want to add that the "See equation 11.1" is on page 299 and is called "(26.1)" for some reason. Also, for the "See equation 11.5" is on page 300 and is called "(26.5)".

  1. The standard free energy change for the formation of two moles of $\ce{H2O(l)}$ in a strong acid-strong base neutralization reaction at $25\ \mathrm{^\circ C}$ is $-79.9\ \mathrm{kJ}$.

$$\ce{H3O+(aq) + OH- (aq) -> 2 H2O(l)}$$

$$\Delta G^\circ = -79.9\ \mathrm{kJ}$$

a. Calculate the equilibrium constant for the reaction. See equation 11.1.

$\Delta G^\circ = -RT\ln K$

$\Delta G^\circ = -79.9\ \mathrm{kJ}$

$R = 8.314 \times 10^{-3}\ \mathrm{kJ/mol\cdot K} = 0.008314\ \mathrm{kJ\cdot mol^{-1}\ K^{-1}}$

$T = 25\ \mathrm{^\circ C} = 298\ \mathrm K$

$-79.9\ \mathrm{kJ} = -\left(0.008314\ \mathrm{kJ\cdot mol^{-1}\ K^{-1}}\right)\left(298\ \mathrm K\right)\ln K$

$K = 1.013 \times 10^{14}$

The equilibrium constant calculated for the reaction is $K = 1.013 \times 10^{14}$

b. Explain the chemical significance of the calculated equilibrium constant for the neutralization reaction.

The explanation for the chemical significance of the calculated equilibrium constant for the neutralization reaction is that its the ratio of the concentrations of the products to the reactants at which the reaction reaches equilibrium. The significance of a larger number is that it has a relatively large amount product.

c. The standard enthalpy change, $\Delta H^\circ$, for a strong acid-strong base reaction is $-57.8\ \mathrm{kJ}$. Determine the standard entropy change, $\Delta S^\circ$, for the reaction at $25\ \mathrm{^\circ C}$. See equation 11.5.

$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$

$\Delta G^\circ = -79.9\ \mathrm{kJ}$

$\Delta H^\circ = -57.8\ \mathrm{kJ}$

$T = 25\ \mathrm{^\circ C} = 298\ \mathrm K$

$-79.9\ \mathrm{kJ} = -57.8\ \mathrm{kJ}-298K\Delta S^\circ$

$\Delta S^\circ = 0.0742\ \mathrm{kJ}$

The standard entropy change, $\Delta S^\circ$, for the reaction at $25\ \mathrm{^\circ C}$ determined is $\Delta S^\circ$ = 0.0742 kJ

d. Explain the chemical significance of the calculated entropy change for the neutralization reaction.

The explanation for the chemical significance of the calculated entropy change for the neutralization reaction is that an increase in the number of microstates results in the calculated change in entropy to be positive.

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    $\begingroup$ Well, just plug in the $\Delta\rm G^0$ and get the $\rm K$. Also, think of a more descriptive title, because you know, every single question on this site is a chemistry question. $\endgroup$ – Ivan Neretin Nov 23 '15 at 18:17
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    $\begingroup$ This seems like a homework question. We ‎have a policy which states that you should show your thoughts and/or efforts into solving the ‎problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this ‎question may get closed. Additionally, everything you need to do the question is given to you, you just need to plug in the numbers. What exactly about this is giving you difficulty?? $\endgroup$ – bon Nov 23 '15 at 18:34
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    $\begingroup$ This question basically boils down to what is the inverse function of lnx. $\endgroup$ – bon Nov 23 '15 at 22:01
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    $\begingroup$ Your problem is not with thermodynamics, chemistry, or physics. It is with mathematics, algebra in particular. You need to review logarithms. $\endgroup$ – Chet Miller Nov 24 '15 at 2:19
  • $\begingroup$ Your enthalpy should be in $\mathrm{kJ/K}$. :) $\endgroup$ – timaeus222 Nov 25 '15 at 7:59
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I assume that your math for parts A and C are correct.

Here is a hint for part B. Which side of the equilibrium is favored?

Also for part D, having a positive $\Delta S$ only implies that there is an increase in the number of microstates in the system and tell you nothing about the spontaneity.

Also, the inverse of $\ln x$ is $\mathrm{e}^x$.

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