A+3BgivesP and M gives 2Q+R if these reactions occur simultaneously in a reactor such that temperature doesnot change.if rate of disappearance of B is (y) Msec^-1 then rate of formation ofQ is

To solve the problem, we need to analyze the two reactions occurring simultaneously in the reactor. We have the reactions A + 3B → P and M → 2Q + R. Given that the rate of disappearance of B is y Msec^-1, we want to find the rate of formation of Q.

Understanding Reaction Rates

In chemical kinetics, the rate of a reaction is defined as the change in concentration of a reactant or product per unit time. For the reactions provided, we can express the rates in terms of the stoichiometry of the reactions.

Analyzing the First Reaction

For the reaction A + 3B → P, the stoichiometry indicates that for every 3 moles of B that disappear, 1 mole of A reacts to form 1 mole of P. Therefore, the rate of disappearance of B can be expressed as:

• Rate of disappearance of B = -1/3 * (d[B]/dt)

Since we know the rate of disappearance of B is y Msec^-1, we can write:

• -1/3 * (d[B]/dt) = y

This implies that:

• d[B]/dt = -3y

Examining the Second Reaction

Now, let's look at the second reaction, M → 2Q + R. The stoichiometry here tells us that for every mole of M that reacts, 2 moles of Q are produced. Therefore, we can express the rate of formation of Q as:

• Rate of formation of Q = 2 * (d[M]/dt)

Relating the Two Reactions

Since the reactions occur simultaneously and the temperature remains constant, we can relate the rates of the two reactions. The rate at which B disappears is directly related to the rate at which M reacts, as both reactions are happening at the same time. We can assume that the consumption of B is linked to the consumption of M based on their stoichiometric coefficients.

Establishing the Relationship

From the first reaction, we know that 3 moles of B are consumed for every mole of A that reacts. If we denote the rate of reaction of M as z Msec^-1, we can relate the two rates as follows:

• 3 * z = -d[B]/dt = 3y

From this, we can conclude that:

• z = y

Calculating the Rate of Formation of Q

Now substituting z into the rate of formation of Q:

• Rate of formation of Q = 2 * z = 2 * y

Final Result

Therefore, the rate of formation of Q is:

• 2y Msec^-1

This means that for every unit of time, the formation of Q is twice the rate at which B is disappearing. This relationship showcases how stoichiometry plays a crucial role in understanding reaction rates in simultaneous reactions.