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The question due to which my doubt arose $\ce{->}$

The following system is in equilibrium: $\ce{SO2Cl2 + {Heat} -> SO2 +Cl2}$

What will happen to the temperature of the system initially if some $\ce{Cl2}$ is added into it at constant volume ?

My attempt at the question : Since if we add more $\ce{Cl2}$, concentration of products will increase which will in turn shift the equilibrium in backward direction, which will in turn release more heat. And since the heat is being released to the surrounding, temperature of the system should decrease.

Though according to my textbook the answer is temperature of the system will increase.

So please do explain where my understanding is wrong.

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  • $\begingroup$ Is the reaction exothermic or endothermic, that is, what is the sign of $\Delta T$? $\endgroup$
    – Zhe
    Mar 25, 2017 at 21:03
  • $\begingroup$ @zhe reaction is endothermic as heat is being supplied $\endgroup$
    – parth_07
    Mar 25, 2017 at 21:49
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    $\begingroup$ The system is in thermal equilibrium with the surroundings. If the temperature of the surroundings increase, then the temperature of the system will increase... $\endgroup$
    – Zhe
    Mar 25, 2017 at 22:04
  • $\begingroup$ @zhe Can you please explain your reply in detail . If the system is in thermal equilibrium then the temperature of both system and surrounding would always be same ? $\endgroup$
    – parth_07
    Mar 25, 2017 at 22:10
  • $\begingroup$ And how do we know that the system is in thermal equilibrium ? Are all chemical reactions in equilibrium are also assumed to be in thermal equilibrium . $\endgroup$
    – parth_07
    Mar 25, 2017 at 22:12

2 Answers 2

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I think this comes down to what constitutes the system. If we think of the system as the reactants and products in a container, releasing heat will increase the temperature of the system.

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  • $\begingroup$ Okay this do seems right to me , so if this is the case than my logic (my attempt in the question ) is correct if we consider only reactants as the system ? $\endgroup$
    – parth_07
    Mar 25, 2017 at 22:18
  • $\begingroup$ I would think you would want to consider the mixture of all the gases to be the system, but in that case your logic is sound. $\endgroup$
    – Tyberius ♦
    Mar 25, 2017 at 22:23
  • $\begingroup$ one more query on the same question .If we think of the system as the reactants and products in a container ,then releasing heat would actually just be transfer of heat from the system to the system .Shouldn't the temperature remain same in this case ? $\endgroup$
    – parth_07
    Mar 25, 2017 at 23:01
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    $\begingroup$ The temperature of a system, to describe it somewhat imprecisely, is related to the kinetic energy of the various particles in the system. The heat released when the reaction runs in reverse comes from the energy released in bond formation, which then becomes kinetic energy, thus increasing the temperature. Similarly, when the reaction goes forward, the energy needed to break bonds is extracted from the kinetic energy of the system, leading to a decrease in temperature. $\endgroup$
    – Tyberius ♦
    Mar 25, 2017 at 23:13
  • $\begingroup$ Oh now I got this , thanks for explaining in detail . $\endgroup$
    – parth_07
    Mar 25, 2017 at 23:15
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This really a comment but too long to fit. The question does not specify if the container holding the gas is to be treated isothermally or adiabatically. However, by inference since the question asks about temperature changes then it cannot be isothermal, i.e. it must be that no heat is gained or lost to the 'world' outside the container otherwise the question does not make sense. It is also implicitly assumed that the added gas is at the same temperature as the gases currently there.

The fact that the volume is fixed means that there is no work done against any external pressure, so all the heat generated by the reaction moving towards reactants heats up the gases.
( Note also that in thermodynamics we always move from one equilibrium state to another equilibrium state; time is never considered.)

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