In the first step of the reaction, which compound is the limiting reagent: sodium hydroxide or the nucleophile

1 answer:

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Since the nucleophile is the actual attacking molecule or molecule that starts the reaction and allows for further steps in the mechanism to occur, it is the limiting reagent, as based on the amount of the nucleophile you have, the reaction will tend to proceed until you run out. The excess would be the sodium hydroxide, it is union part of the solution.

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Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$