b - the repulsion of the protons is decreasing.
Answer : The entropy change of reaction for 1.62 moles of 
reacts at standard condition is 217.68 J/K
Explanation :
The given balanced reaction is,
The expression used for entropy change of reaction 
is:
![\Delta S^o=[n_{Br_2}\times \Delta S_f^0_{(Br_2)}+n_{F_2}\times \Delta S_f^0_{(F_2)}]-[n_{BrF_3}\times \Delta S_f^0_{(BrF_3)}]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5Bn_%7BBr_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28Br_2%29%7D%2Bn_%7BF_2%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28F_2%29%7D%5D-%5Bn_%7BBrF_3%7D%5Ctimes%20%5CDelta%20S_f%5E0_%7B%28BrF_3%29%7D%5D)
where,
= entropy change of reaction = ?
n = number of moles
= standard entropy of formation
= 245.463 J/mol.K
= 202.78 J/mol.K
= 292.53 J/mol.K
Now put all the given values in this expression, we get:
![\Delta S^o=[1mole\times (245.463J/K.mole)+3mole\times (202.78J/K.mole)}]-[2mole\times (292.53J/K.mole)]](https://tex.z-dn.net/?f=%5CDelta%20S%5Eo%3D%5B1mole%5Ctimes%20%28245.463J%2FK.mole%29%2B3mole%5Ctimes%20%28202.78J%2FK.mole%29%7D%5D-%5B2mole%5Ctimes%20%28292.53J%2FK.mole%29%5D)
Now we have to calculate the entropy change of reaction for 1.62 moles of reacts at standard condition.
From the reaction we conclude that,
As, 2 moles of has entropy change = 268.74 J/K
So, 1.62 moles of has entropy change = 
Therefore, the entropy change of reaction for 1.62 moles of reacts at standard condition is 217.68 J/K
Answer:
0.54 mole
Explanation:
CH3COOH CH3CH2OH CH3COOCH2CH3 H2O
Initial concentration 1.0 mole 1.0 mole 0 mole 1.0mol
Change - x - x + x + x
Equilibrium (1.0 - x) (1.0 - x) x (1.0 + x)
K = [CH3COOCH2CH3]*[H2O]/[CH3COOH]*[CH3CH2OH]
x*(1.0+x)/(1.0-x)(1.0-x) = 4.0
x+x²=4*(1-x)²
x+x² = 4(1² - 2x + x²)
x + x² = 4 - 8x + 4x²
4 - 8x + 4x²- x² - x= 0
3x² - 9x + 4 = 0
x=2.5 , x=0.54
2.5 mole of acid cannot be esterified, because there is only 1.0 mole of acid,
so answer is 0.54 mole.
The way to know this is by using the next procedure:
<span>n = c x v
0.665 = 1.15 x v
v = 0.578L = 578mL QED
This is going to help you indeed. Hope its good for you
</span>
