Question

Two gases A and B which react according to the equation

$aA_{(g)} + bB_{(g)} \longrightarrow cC_{(g)} + dD_{(g)}$

to give two gases C and D are taken (amount not known) in an Eudiometer tube (operating constant Pressure and temperature) to cause the above.

If on causing the reaction there is no volume change observed then which of the following state is/are correct.

• A. (a + b) = (c + d)
• B. average molecular mass may increase or decrease if either of A or B is present in limited am
• C. Vapour Density of the mixture will remain same throughout the course of reaction.
• D. Total moles of all the component of mixture will change.

Answer 1

Answer: (A), (C)

A, C

Answer 2

The reaction given is:

$aA_{(g)} + bB_{(g)} \longrightarrow cC_{(g)} + dD_{(g)}$

The reaction is carried out in an Eudiometer tube at constant pressure (P) and constant temperature (T). It is observed that there is no volume change during the reaction. This means the initial volume of the gaseous mixture is equal to the final volume of the gaseous mixture. According to Avogadro's Law, at constant P and T, the volume of a gas is directly proportional to the number of moles of the gas ($V \propto n$). Therefore, if there is no volume change, it implies that the total number of moles of gaseous components remains constant throughout the reaction.

$n_{initial} = n_{final}$

Let's analyze each statement:

(A) (a + b) = (c + d)

Let $n_A^0$ and $n_B^0$ be the initial moles of gases A and B, respectively. Initial total moles of gas = $n_A^0 + n_B^0$. Let's assume the reaction proceeds to an extent where $x$ moles of A react. Then, moles of B reacted = $(b/a)x$. Moles of C formed = $(c/a)x$. Moles of D formed = $(d/a)x$. If A is the limiting reactant, then $x = n_A^0$.

Moles of A remaining = 0

Moles of B remaining = $n_B^0 - (b/a)n_A^0$

Moles of C formed = $(c/a)n_A^0$

Moles of D formed = $(d/a)n_A^0$

Final total moles of gas ($n_{final}$) = (moles of unreacted B) + (moles of C) + (moles of D)

$n_{final} = (n_B^0 - (b/a)n_A^0) + (c/a)n_A^0 + (d/a)n_A^0$

$n_{final} = n_B^0 + \frac{n_A^0}{a} (c + d - b)$

Since $n_{initial} = n_{final}$:

$n_A^0 + n_B^0 = n_B^0 + \frac{n_A^0}{a} (c + d - b)$

$n_A^0 = \frac{n_A^0}{a} (c + d - b)$

Assuming $n_A^0 \neq 0$, we can divide by $n_A^0$:

$1 = \frac{1}{a} (c + d - b)$

$a = c + d - b$

Rearranging, we get: $a + b = c + d$. This statement is correct.

(B) average molecular mass may increase or decrease if either of A or B is present in limited amount.

The average molecular mass ($\bar{M}$) of a gaseous mixture is defined as:

$\bar{M} = \frac{\text{Total mass of the mixture}}{\text{Total moles of the mixture}}$

According to the law of conservation of mass, the total mass of the system remains constant throughout the chemical reaction. From the analysis of statement (A), we concluded that the total number of moles of gaseous components also remains constant ($n_{initial} = n_{final}$). Since both the total mass of the mixture and the total moles of the mixture remain constant, the average molecular mass of the mixture must also remain constant. Therefore, this statement is incorrect.

(C) Vapour Density of the mixture will remain same throughout the course of reaction.

Vapour Density (VD) is related to the average molecular mass ($\bar{M}$) by the formula:

$VD = \frac{\bar{M}}{2}$ (when relative to hydrogen, $M_{H_2}=2$)

Since we established that the average molecular mass ($\bar{M}$) remains constant, the Vapour Density of the mixture will also remain constant. This statement is correct.

(D) Total moles of all the component of mixture will change.

As derived from the condition of no volume change at constant P and T, the total moles of gaseous components in the mixture remain constant ($n_{initial} = n_{final}$). Therefore, this statement is incorrect.

Based on the analysis, statements (A) and (C) are correct.