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Q6.25(c):
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (c) $3Fe (s) + 4H_2O (g) \rightleftharpoons Fe_3O_4 (s) + 4H_2 (g)$

Solution :

Step 1: Analyzing the Balanced Chemical Equation

The given equilibrium reaction is:

$3Fe(s) + 4H_2O(g) \rightleftharpoons Fe_3O_4(s) + 4H_2(g)$

To determine the effect of pressure change on the equilibrium, we must identify the state of matter for each participant. In gaseous equilibria, the shift is dictated by the total number of moles of gaseous species on each side of the equation [Per Le Chatelier's Principle].

Step 2: Counting Gaseous Moles ($n_g$)

We classify the species by their phase ($s = solid$, $g = gas$):

  • Reactants: $3Fe(s)$ and $4H_2O(g)$. Total moles of gas ($n_{g, reactants}$) = 4.
  • Products: $Fe_3O_4(s)$ and $4H_2(g)$. Total moles of gas ($n_{g, products}$) = 4.

Step 3: Calculating Change in Gaseous Moles ($\Delta n_g$)

The change in the number of gaseous moles is calculated as follows:

$\Delta n_g = \sum n_{g, products} - \sum n_{g, reactants}$

$\Delta n_g = 4 - 4 = 0$

Step 4: Evaluating the Effect of Pressure/Volume Change

Le Chatelier's Principle states that when pressure is decreased (by increasing volume) in a system at equilibrium, the system will shift to the side with a greater number of gaseous moles to counteract the pressure drop. Conversely, if pressure is increased, it shifts toward the side with fewer gaseous moles.

In this specific reaction, since $\Delta n_g = 0$:

  • The number of moles of gaseous products does not depend on the total pressure of the system.
  • A change in volume/pressure does not create a concentration gradient that would favor the forward or backward reaction [Per the Law of Mass Action].
$\Delta n_g = 0$ Reactant Gas: 4 mol Product Gas: 4 mol

Step 5: Conclusion

Because the number of moles of gaseous reactants is equal to the number of moles of gaseous products, the equilibrium position remains unaffected by changes in pressure or volume.

Final Solution: The number of moles of reaction products remains the same.


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