When equal volumes of ethanol and water are mixed to form a solution, the entropy of the resulting solution increases. This is because ethanol and water are both polar substances, making them miscible. The solubility of both substances is high because they have similar attractions due to the presence of hydrogen bonding. This means that molecules of water and ethanol freely move with each other, therefore increasing the dispersal, randomness, and disorder (entropy) of the system.
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Answer:
The equilibrium will move in forward direction.
Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On addition of base at the equilibrium, the hydroxide ions of the base will neutralize the hydrogen ions and lowering in concentration of hydrogen ion will be observed.
So, on lowering of concentration of hydrogen ions the equilibrium will move in direction in accordance to Le Chatelier’s Principle .The equilibrium will move in forward direction.
For a general reaction,

General expression for rate law will be:
![r=k[A]^{a}[B]^{b}](https://tex.z-dn.net/?f=r%3Dk%5BA%5D%5E%7Ba%7D%5BB%5D%5E%7Bb%7D)
Here, r is rate of the reaction, k is rate constant, a is order with respect to reactant A and b is order with respect to reactant B.
The reaction is first order with respect to
, second order with respect to
and zero order with respect to
.
According to above information, expression for rate law will be:
![r=k[BrO_{3}^{-}]^{1}[Br^{-}]^{2}[H^{+}]^{0}](https://tex.z-dn.net/?f=r%3Dk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5E%7B1%7D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%5BH%5E%7B%2B%7D%5D%5E%7B0%7D)
Or,
...... (1)
- When concentration of
get doubled, rate of the reaction becomes,
...... (2)
Dividing (2) by (1)
![\frac{r^{'}}{r}=\frac{2k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=2](https://tex.z-dn.net/?f=%5Cfrac%7Br%5E%7B%27%7D%7D%7Br%7D%3D%5Cfrac%7B2k%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%7Bk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%3D2)
Or,

Thus, rate of the reaction also get doubled.
- When the concentration of
is halved, the rate of reaction becomes
Or,
...... (3)
Dividing (3) by (1)
![\frac{r^{"}}{r}=\frac{1/4k[BrO_{3}^{-}][Br^{-}]^{2}}{k[BrO_{3}^{-}][Br^{-}]^{2}}=\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7Br%5E%7B%22%7D%7D%7Br%7D%3D%5Cfrac%7B1%2F4k%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%7Bk%5BBrO_%7B3%7D%5E%7B-%7D%5D%5BBr%5E%7B-%7D%5D%5E%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B4%7D)
Or,

Thus, rate of reaction becomes 1/4th of the initial rate.
- When the concentration of
is tripled:
Since, the rate expression does not have concentration of
, it is independent of it. Thus, any change in the concentration will not affect the rate of reaction and rate of reaction remains the same as in equation (1).