Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:


The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)


The work:

If it is an ideal gas:


Solving:

Replacing:


Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
Question is incomplete, complete question is;
A 34.8 mL solution of
(aq) of an unknown concentration was titrated with 0.15 M of NaOH(aq).

If it takes 20.4 mL of NaOH(aq) to reach the equivalence point of the titration, what is the molarity of
? For your answer, only type in the numerical value with two significant figures. Do NOT include the unit.
Answer:
0.044 M is the molarity of
(aq).
Explanation:
The reaction taking place here is in between acid and base which means that it is a neutralization reaction .
To calculate the concentration of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of acid which is 
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

0.044 M is the molarity of
(aq).
Answer: The value of
for this reaction is 250000.
Explanation:
The given equation is as follows.

... (1)
... (2)
To balance the atoms, multiply equation (2) by 3. Hence, the equation (2) can be re-written as follows.
... (3)
Now, subtract equation (1) from equation (3). So, the equation formed will be as follows.

This equation can also be re-written as follows.

This equation is similar to the equilibrium equation given to us.
Therefore, during this subtraction the equation constants get divided as follows.
Thus, we can conclude that the value of
for this reaction is 250000.