Amole of x reacts at a constant pressure of 43.0 atm via the reaction x(g)+4y(g)→2z(g), δh∘=−75.0 kj before the reaction, the volume of the gaseous mixture was 5.00 l. after the reaction, the volume was 2.00 l. calculate the value of the total energy change, δe, in kilojoules.

The volume fraction in gasses is equal to the mole fraction.
In a reaction mixture, the molar ratio of X : Y = 1:4
Fraction of X = 1/5
Fraction of Y = 4/5
So in the 5 L reaction mix,
1 L of X and 4 L of Y
The products contain 2 L of Z
This satisfies the equation completely so we check the enthalpy of this reaction:
-75 kilojoules

The answer to your question really depends on three things: how much water you have, how cold it is to start out, and how cold the things around it are. water actually freezes when it gets to 32 degrees fahrenheit (0 degrees celsius), but the time it takes to get there may be different.  let's start with the first. if you take two glasses, and fill one with a tiny bit of water, and the other about halfway, then put them both in the freezer, the one with less water will freeze first (you can try this at home, but i recommend using plastic cups and not glass ones).  now let's move on to the second part. let's say you have two glasses, and you fill one with really cold water that has been in the refrigerator, and the other with really hot water from the sink. if you put both of them in the freezer, the one that started out colder will freeze first.  for the third part, let's imagine that you have two glasses with the same amount of water in them, and the water is at the same temperature. imagine putting one outside on a really really cold day in georgia, and having a friend in alaska put one outside on the same day. since it would be so much colder in alaska, the glass of water there would freeze before yours.  so, if you took a tiny bit of really cold water in a glass, and put it outside on a cold day in alaska, it would freeze a lot faster than a big glass of hot water outside on a cold day in georgia. 

The bergeron process.