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Consider the following equilibrium reaction: $$\ce{2NOCl_{(g)} <=> 2NO_{(g)} + Cl_{2(g)}} \\ \Delta{H}>0$$ My

$$\ce{2 NOCl (g) <=> 2 NO (g) + Cl2 (g)} \qquad \Delta H > 0$$

My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta{H}$$\Delta H$ is positive, we can think of the reaction as: $$\ce{2NOCl_{(g)} + heat <=> 2NO_{(g)} + Cl_{2(g)}}$$ And

$$\ce{2 NOCl (g) + \text{heat} <=> 2 NO (g) + Cl2 (g)}$$

And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have $heat$heat in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: Pressurepressure. 

While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one. 

Is my reasoning true? If it is, how can iI conclude the side determining job?

Consider the following equilibrium reaction: $$\ce{2NOCl_{(g)} <=> 2NO_{(g)} + Cl_{2(g)}} \\ \Delta{H}>0$$ My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta{H}$ is positive, we can think of the reaction as: $$\ce{2NOCl_{(g)} + heat <=> 2NO_{(g)} + Cl_{2(g)}}$$ And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have $heat$ in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: Pressure. While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one. Is my reasoning true? If it is, how can i conclude the side determining job?

Consider the following equilibrium reaction:

$$\ce{2 NOCl (g) <=> 2 NO (g) + Cl2 (g)} \qquad \Delta H > 0$$

My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta H$ is positive, we can think of the reaction as:

$$\ce{2 NOCl (g) + \text{heat} <=> 2 NO (g) + Cl2 (g)}$$

And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have heat in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: pressure. 

While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one. 

Is my reasoning true? If it is, how can I conclude the side determining job?

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user43236
user43236

Does temperature change pressure in isochoric system?

Consider the following equilibrium reaction: $$\ce{2NOCl_{(g)} <=> 2NO_{(g)} + Cl_{2(g)}} \\ \Delta{H}>0$$ My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta{H}$ is positive, we can think of the reaction as: $$\ce{2NOCl_{(g)} + heat <=> 2NO_{(g)} + Cl_{2(g)}}$$ And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have $heat$ in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: Pressure. While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one. Is my reasoning true? If it is, how can i conclude the side determining job?