Question

The reaction $3A + B \rightarrow 2C$ follows a rate law where the rate of disappearance of A is half the rate of appearance of C, and experimental data shows:

• When $[A]$ is doubled (holding $[B]$ constant), the rate increases by a factor of 8.
• When $[B]$ is tripled (holding $[A]$ constant), the rate increases by a factor of 9.

Given $[A]_0 = 0.6 \, \text{M}$, $[B]_0 = 0.4 \, \text{M}$, and the rate constant $k = 0.03 \, \text{M}^{-3} \text{s}^{-1}$ at 400 K, calculate:

1. The time (in seconds) for $[B]$ to drop to 0.1 M.
2. The instantaneous rate of formation of C at the moment when $[A] = 0.3 \, \text{M}$.
3. The half-life of A if the reaction were run with $[A]_0 = 0.8 \, \text{M}$ and $[B]_0 = 0.2 \, \text{M}$.

Answer 1

1. It is impossible for $[B]$ to drop to 0.1 M under the given initial conditions.
2. $1.458 \times 10^{-5} \, M s^{-1}$
3. The calculation requires solving a complex integral, which is beyond the scope of typical analytical methods for JEE/NEET.

Answer 2

The problem contains multiple inconsistencies, which are common in poorly formulated questions. We will address these inconsistencies and make necessary assumptions to solve the problem.

Inconsistencies and Assumptions:

1. Rate Law and Stoichiometry Conflict: The problem states "the rate of disappearance of A is half the rate of appearance of C", i.e., $-d[A]/dt = 1/2 d[C]/dt$. However, from the stoichiometry of the reaction $3A + B \rightarrow 2C$, the relationship is $-1/3 d[A]/dt = +1/2 d[C]/dt$, which simplifies to $-d[A]/dt = 3/2 d[C]/dt$. These two statements contradict each other. In chemical kinetics, the stoichiometric coefficients define the relative rates of consumption/formation. Therefore, we will disregard the statement -d[A]/dt = 1/2 d[C]/dt and use the standard stoichiometric relationships: $\text{Rate} = -\frac{1}{3}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = +\frac{1}{2}\frac{d[C]}{dt}$

2. Rate Constant Units Conflict:

   • From the experimental data:

     • When $[A]$ is doubled, rate increases by a factor of 8. This means $(2)^x = 8 \implies x=3$.
     • When $[B]$ is tripled, rate increases by a factor of 9. This means $(3)^y = 9 \implies y=2$.

   • So, the rate law is $\text{Rate} = k[A]^3[B]^2$. The overall order of the reaction is $n = x+y = 3+2 = 5$.

   • For an $n$-th order reaction, the units of the rate constant $k$ are $(\text{concentration})^{1-n} (\text{time})^{-1}$. For $n=5$, the units should be $M^{1-5} s^{-1} = M^{-4} s^{-1}$.

   • The problem states $k = 0.03 \, M^{-3} s^{-1}$. This unit corresponds to a 4th-order reaction.

   • Since the experimental data for determining orders is unambiguous, we will assume the numerical value of $k$ is correct, but its unit is misstated. We will use $k = 0.03 \, M^{-4} s^{-1}$ consistent with the derived 5th order.

3. Impossible Concentration Drop for [B]:

   • Initial concentrations: $[A]_0 = 0.6 \, M$, $[B]_0 = 0.4 \, M$.
   • The question asks for the time for $[B]$ to drop to $0.1 \, M$. This means $\Delta[B] = 0.4 - 0.1 = 0.3 \, M$.
   • From stoichiometry $3A + B \rightarrow 2C$, if $0.3 \, M$ of B reacts, then $3 \times 0.3 = 0.9 \, M$ of A would be consumed.
   • However, we only have $[A]_0 = 0.6 \, M$. Since $[A]$ is the limiting reactant ($0.6/3 = 0.2$ vs $0.4/1 = 0.4$), the reaction will stop when A is completely consumed.
   • When A is consumed, $\Delta[A] = 0.6 \, M$. This implies $\Delta[B] = 0.6/3 = 0.2 \, M$.
   • So, $[B]$ will drop to a minimum of $[B]_0 - 0.2 = 0.4 - 0.2 = 0.2 \, M$. It cannot drop to $0.1 \, M$.
   • Conclusion for Part 1: It is physically impossible for $[B]$ to drop to $0.1 \, M$ under the given initial conditions and stoichiometry.

Given these significant flaws, we will proceed to solve parts 2 and 3, and for part 1, we will state that it's impossible. If we were forced to provide a numerical answer for part 1, we would have to assume a different initial concentration for A (e.g., A in large excess) or that the question implies a pseudo-order condition, which is not stated. For the purpose of a standard exam, stating impossibility is the correct response for part 1.

Refined Rate Law: $\text{Rate} = k[A]^3[B]^2$ where $k = 0.03 \, M^{-4} s^{-1}$.

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1. The time (in seconds) for $[B]$ to drop to 0.1 M.

As explained above, based on the initial concentrations $[A]_0 = 0.6 \, M$ and $[B]_0 = 0.4 \, M$, A is the limiting reactant. For $[B]$ to drop by $0.3 \, M$ (from $0.4 \, M$ to $0.1 \, M$), $0.9 \, M$ of A would be required. Since only $0.6 \, M$ of A is available, the reaction will cease when A is consumed. At that point, $[A] = 0 \, M$, and $[B]$ will have dropped by $0.6/3 = 0.2 \, M$, reaching $[B] = 0.4 - 0.2 = 0.2 \, M$. Therefore, it is impossible for $[B]$ to drop to $0.1 \, M$.

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2. The instantaneous rate of formation of C at the moment when $[A] = 0.3 \, M$.

First, find the concentration of B when $[A] = 0.3 \, M$. Initial $[A]_0 = 0.6 \, M$, $[B]_0 = 0.4 \, M$. Change in $[A]$, $\Delta[A] = 0.6 - 0.3 = 0.3 \, M$. From stoichiometry $3A + B \rightarrow 2C$, for every 3 moles of A consumed, 1 mole of B is consumed. So, $\Delta[B] = \Delta[A]/3 = 0.3 \, M / 3 = 0.1 \, M$. Concentration of B at this moment, $[B] = [B]_0 - \Delta[B] = 0.4 - 0.1 = 0.3 \, M$.

Now, calculate the instantaneous reaction rate: $\text{Rate} = k[A]^3[B]^2$ $\text{Rate} = (0.03 \, M^{-4} s^{-1})(0.3 \, M)^3(0.3 \, M)^2$ $\text{Rate} = (0.03)(0.027)(0.009) \, M s^{-1}$ $\text{Rate} = 0.03 \times 0.000243 \, M s^{-1}$ $\text{Rate} = 7.29 \times 10^{-6} \, M s^{-1}$

The instantaneous rate of formation of C, $d[C]/dt$, is related to the reaction rate by stoichiometry: $\text{Rate} = +\frac{1}{2}\frac{d[C]}{dt}$ So, $\frac{d[C]}{dt} = 2 \times \text{Rate}$ $\frac{d[C]}{dt} = 2 \times (7.29 \times 10^{-6} \, M s^{-1})$ $\frac{d[C]}{dt} = 1.458 \times 10^{-5} \, M s^{-1}$

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3. The half-life of A if the reaction were run with $[A]_0 = 0.8 \, M$ and $[B]_0 = 0.2 \, M$.

First, identify the limiting reactant for these new initial conditions: For A: $0.8 \, M / 3 = 0.267$ For B: $0.2 \, M / 1 = 0.2$ Since $0.2 < 0.267$, B is the limiting reactant.

When B is completely consumed, $\Delta[B] = 0.2 \, M$. From stoichiometry, $\Delta[A] = 3 \times \Delta[B] = 3 \times 0.2 = 0.6 \, M$. So, $[A]$ will drop from $0.8 \, M$ to $0.8 - 0.6 = 0.2 \, M$. The reaction stops when B is consumed. The concept of "half-life of A" usually implies the time for $[A]$ to reach $[A]_0/2$, which is $0.8/2 = 0.4 \, M$. For $[A]$ to drop to $0.4 \, M$, $\Delta[A] = 0.8 - 0.4 = 0.4 \, M$. This requires $\Delta[B] = 0.4/3 = 0.133 \, M$. Since $[B]_0 = 0.2 \, M$, and $0.133 \, M < 0.2 \, M$, there is enough B for A to reach its half-life. So, half-life of A is a meaningful concept here.

Since neither reactant is in large excess, this is not a pseudo-order reaction. We need to use the integrated rate law considering both reactants. Let $x$ be the amount of B reacted at time $t$. $[B]_t = [B]_0 - x$ $[A]_t = [A]_0 - 3x$

The rate law is $-\frac{d[B]}{dt} = k[A]^3[B]^2$. Substitute $[A]$ and $[B]$ in terms of $x$: $-\frac{d([B]_0 - x)}{dt} = k([A]_0 - 3x)^3([B]_0 - x)^2$ $\frac{dx}{dt} = k([A]_0 - 3x)^3([B]_0 - x)^2$ To find the half-life of A, we need to find the time $t_{1/2}$ when $[A]_t = [A]_0/2$. $[A]_0 - 3x = [A]_0/2$ $3x = [A]_0/2$ $x = [A]_0/6$ Given $[A]_0 = 0.8 \, M$, $x = 0.8/6 = 0.4/3 \, M$.

So, we need to solve the integral: $\int_0^{0.4/3} \frac{dx}{([A]_0 - 3x)^3([B]_0 - x)^2} = \int_0^{t_{1/2}} k \, dt = k t_{1/2}$ Substituting $[A]_0 = 0.8$ and $[B]_0 = 0.2$: $\int_0^{0.4/3} \frac{dx}{(0.8 - 3x)^3(0.2 - x)^2} = 0.03 \, t_{1/2}$ This integral is very complex and cannot be solved analytically using methods typically taught in JEE/NEET. This suggests that either the problem expects a numerical method (which is unlikely), or there is an implicit assumption that simplifies the problem, or the problem is flawed in its expectation of a simple analytical solution.

In typical competitive exams, if such a complex integral arises, it usually implies that one of the reactants is in very large excess, allowing for a pseudo-order approximation. However, here $[A]_0 = 0.8 \, M$ and $[B]_0 = 0.2 \, M$ are comparable, so no such approximation is valid.

Given the constraints of a JEE/NEET level problem, a direct analytical solution for such a complex integral is not expected. Therefore, we must conclude that this part of the question is also beyond the scope of analytical solution without further simplifying assumptions not provided.

If forced to provide a numerical answer, one would need computational tools for integration. Without that, it's not feasible.

Final Answer Summary:

1. Time for $[B]$ to drop to 0.1 M: It is impossible for $[B]$ to drop to 0.1 M under the given initial conditions because A is the limiting reactant and would be consumed before $[B]$ reaches 0.1 M.
2. Instantaneous rate of formation of C at $[A] = 0.3 \, M$: $1.458 \times 10^{-5} \, M s^{-1}$
3. Half-life of A with new initial conditions: The calculation requires solving a complex integral for a multi-reactant, non-pseudo-order reaction, which is beyond the scope of typical analytical methods for JEE/NEET.

Since the question asks to "calculate", and provides specific values, it implies a calculable answer. The impossibility of the first part and the complexity of the third part's integral suggest significant flaws in the question design. However, for the second part, a direct calculation is possible.