Answer:
Pleae see below
Explanation:
According to the first law of thermodynamics the change in energy, ΔE , in a system is equal to:
ΔE = q + w
where q is the heat flow involved, and w is the work done on or by the system.
If q flows out of the system, it is negative.
If q flows in from the surroundings into the system, it is positve.
If work is done by the system, it is negative.
If work is done on thesystem, it is positve.
Wr are told that in this chemical reaction 336 kJ are released, thus
ΔE = -336 kJ
The negative sign arises from the fact that the energy is released, i.e.exothermic reaction.
We are also told that 136 kJ of heat flows out of the system, therefore
q = - 136 kJ
The value of q is negative because heat flows out of the system into the water bath.
So we have the following equation:
ΔE = q+w
-336 kJ = - 136 kJ + w
w = -336 kJ + 136 kJ = - 200 kJ
With this in mind, lets answer our questions:
Is the reaction exothermic or endothermic?
The reaction is exothermic, we know 336 kJ of energy are released as stated in the question.
Does the temperature of the water go up or down.
Since the reaction releases energy, and from the fact that as stated heat flows out of the system, the water temperature goes up.
Does the piston move in or out?
The piston moves out since our work is negative ( - 200 kJ) and work against a constant external pressure, as stated in the question is by definition:
w = - pΔV ⇒ ΔV = Vf - Vi must be positive for the overall work to be
-200 kJ as calculated and the piston therefore has moved out.
Does work is done on (by) ?
Remember the extenal pressure is constant, and the piston has moved out so the gas reaction does work of expansion, and the work is done by the system.
How much work is done on (or) by the gas mixture?
As calculated prevoiusly, the work done by the gas mixture is 200 kJ
.