Ask Question
22 September, 01:31

A mixture of gaseous reactants is put in to a cylinder, where a chemical reaction turns them in to gaseous products. The cylinder has a piston that moves in or out as necessary to keep a constant pressure on the mixture of 1 atm. The cylinder is also submerged in a large insulated water bath.

From previous experiments, this chemical reaction is known to release 244. kJ of energy.

The temperature of the water bath is monitored, and it is determined from this data that 150. kJ of heat flows out of the system during the reaction.

1) is the reaction exothermic or endothermic?

2) Does the temperature of the water go up or down?

3) does the piston move in, out, or neither?

4) does The gas mixture do work, Or is work done on it?

5) how much work is done on (or by) the gas mixture?

+5
Answers (1)
  1. 22 September, 01:50
    0
    1) Exothermic.

    2) It goes up.

    3) It moves out.

    4) It does work.

    5) 94 kJ.

    Explanation:

    1) An endothermic reaction is a reaction that absorbs heat from the surroundings, and an exothermic reaction is a reaction that releases heat to the surroundings. So the reaction placed is exothermic.

    2) Because the heat is flowing to the water, its temperature will go up.

    3) By the first law of the thermodynamics:

    ΔU = Q - W

    Where ΔU is the total energy, Q is the heat, and W is the work. Because the energy and the heat are being released, they are both negative:

    -244 = - 150 - W

    W = 94 kJ

    The work is positive, so it's being doing by the system, it means that the system is expanding, and the piston moves out.

    4) As explained above, the gas mixture (the system) does work.

    5) As shown above, W = 94 kJ
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “A mixture of gaseous reactants is put in to a cylinder, where a chemical reaction turns them in to gaseous products. The cylinder has a ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers