When 2.00 mol of SO2Cl2 is placed in a 2.00-L flask at 303
K, 56%...

Question

When 2.00 mol of SO2Cl2 is placed in a 2.00-L flask at 303
K, 56% of the SO2Cl2 decomposes to SO2 and Cl2 :
SO2Cl21g2 Δ SO21g2 + Cl21g2
(c) According to Le Châtelier’s principle, would the percent of SO2Cl2
that decomposes increase, decrease or stay the same if the
mixture were transferred to a 15.00-L vessel?

Accepted Answer

Hi everyone for this problem. It reads at 298 Calvin a three most sample of sulfur tri oxide is added to a three liter vessel, 45% of sulfur trioxide decomposes into sulfur dioxide and oxygen gas. The reaction is given. If the reaction was carried out in a one liter vessel, will the percent of sulfur trioxide that decomposes increase decrease or remain the same? Okay, So our goal is to answer whether or not will have an increase decrease or no change in the percent of sulfur trioxide that decomposes. So let's just go ahead and rewrite our rewrite our reaction. Okay, so we have two moles of sulfur trioxide at equilibrium and we have two moles of sulfur dioxide and one mole of oxygen gas. So as we can see here, we're dealing with gasses and only gasses in a closed system. The pressure of a gas at a constant temperature and volume depends on the number of particles present in the system. So, looking at our reaction, we have two moles of reactant and three moles of products. Only gasses are affected by changes in pressure and volume. And the shot leers principle tells us that if a stress is put on a system that's at equilibrium, the system will try to adjust to minimize that stress. So we need to take a look here at the relationship between pressure and volume, pressure and volume have an inverse relationship. So if we increase pressure, that means there's going to be a decrease in volume. Okay, so an increase in pressure causes the reaction to shift towards the less moles of gas. Okay. And so our equilibrium will favor the direction with the lesson moles of gas. And by looking at our reaction, we said, that's going to be the reactant. So we said we have two moles of reactant versus three moles of product, so decreasing the volume is going to cause an an increase in reactant. So what that means is we'll have more moles of sulfur tri oxide or the percent of sulfur trioxide decomposed will decrease. So this is going to be our final answer for this problem is that the percent of sulfur trioxide that decomposes will decrease. That's the end of this problem. I hope this was helpful.

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