
- <em>Expansion </em><em>of </em><em>particles</em><em> </em><em>of</em><em> </em><em>substances.</em><em> </em>
- <em>Increase</em><em> </em><em>in </em><em>temperature</em><em>.</em>
- <em>Change</em><em> </em><em>in </em><em>state</em><em>.</em>
- <em>Change</em><em> </em><em>in </em><em>physical</em><em> </em><em>property</em>
- <em>It </em><em>may </em><em>bring</em><em> </em><em>out </em><em>chemical</em><em> </em><em>changes</em><em>.</em>
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Answer:
6.142 moles of NaCl
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
2AlCl3 + 3Na2S —> Al2S3 + 6NaCl
Next, we determine the number of mole in 239.7 g of Na2S. This is illustrated below:
Mass mass of Na2S = 78.048g/mol
Mass of Na2S = 239.7g
Number of mole Na2S =..?
Mole = Mass /Molar Mass
Number of mole Na2S = 239.7/78.048 = 3.071 moles
Finally, we can obtain the number of mole of NaCl produced from the reaction as follow:
From the balanced equation above,
3 moles of Na2S reacted to produce 6 moles of NaCl.
Therefore, 3.071 moles of Na2S will react to produce = (3.071 x 6)/3 = 6.142 moles of NaCl
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)