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Chapter 14: Problem 52

Imagine the following hypothetical reaction, taking place in a sealed, rigid container, to be neither exothermic nor endothermic. $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftarrows \mathrm{C}(g) \quad \Delta H=0$$ Would an increase in temperature favor the forward reaction or the reverse reaction? (Hint: Recall the gas laws.)

Short Answer

Expert verified
An increase in temperature would not favor either the forward or the reverse reaction.

Step by step solution

01

Analyzing the reaction

First of all, understand the given reaction and observe that it consists of one molecule of A and one molecule of B forming one molecule of C. It's also important to note that the reaction is neither exothermic nor endothermic. In other words, ΔH = 0, which means the heat change of the system is zero.
02

Applying the Principle of Le Chatelier

Recall the Le Chatelier's Principle: 'if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.' However, since ΔH = 0, changes in temperature have no effect on the equilibrium.
03

Considering Gas Laws

With reference to the hint, consider the Gas Laws, particularly the fact that increasing temperature increases the pressure. In this specific scenario, however, since there are equal numbers of moles of gas on either side of the equilibrium relationship, the reaction would not shift in either direction due to change in temperature.
04

Conclusion

In conclusion, for this specific reaction, neither the forward nor the reverse reactions would be favored by an increase in temperature.

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