Q.The decomposition of $NH_3$ on platinum surface is zero order reaction. What are the rates of production of $N_2$ and $H_2$ if $k = 2.5\times10^{-4}\ \text{mol}^{-1}\,\text{L}\,\text{s}^{-1}$?

Concept and Intuition

The units of a rate constant reveal the reaction order without needing any other data:

• Zero order: $k$ has units of concentration/time ($mol\,L^{-1}\,s^{-1}$ or similar).
• First order: $k$ has units of (time)$^{-1}$ only.
• Second order: $k$ has units of $(concentration)^{-1}(time)^{-1}$. Here $k = 0.02\ mol\,L^{-1}\,min^{-1}$ has units of concentration/time, so this is unambiguously a zero-order reaction, and the integrated rate law to use is the simple linear one, $[A] = [A]_0 - kt$.

Step-by-Step Solution

1. Identify order from units of $k$: $mol\,L^{-1}\,min^{-1} \Rightarrow$ zero order.
2. Zero-order integrated rate law: $[A] = [A]_0 - kt$.
3. Substitute $[A]_0 = 1.0\ mol\,L^{-1}$, $k=0.02\ mol\,L^{-1}\,min^{-1}$, $t = 20$ min:

4. So the concentration of A after 20 min is $0.6\ mol\,L^{-1}$.

Common Mistakes

• Reflexively applying the first-order formula $[A]=[A]_0e^{-kt}$ (and reaching for the given antilog value) without first checking that the units of $k$ actually indicate zero order.
• Misreading the units of $k$ and assuming first order by default.

ANSWER: B

Concept and Intuition

For a zero-order reaction, rate $= k[A]^0 = k$, a constant that does not depend on the concentration of reactant present. Physically, this happens for the catalytic decomposition of $NH_3$ on a hot metal (Pt) surface at high pressure, where the metal surface is fully saturated with adsorbed $NH_3$ molecules — the reaction rate is then limited only by the fixed number of active catalytic sites, not by how much $NH_3$ is left in the gas phase. As a result the rate does not change as the reaction proceeds (until the surface is no longer saturated).

Step-by-Step Solution

1. Zero-order kinetics: rate $= k$ (a constant), independent of reactant concentration and therefore independent of the time elapsed (as long as the surface stays saturated).
2. At $t = 10$ min, rate $= x$ (given).
3. At $t = 20$ min, since rate does not change with time for a zero-order reaction, rate $= x$ again.

Common Mistakes

• Applying first-order-style reasoning (assuming rate decreases as reactant is consumed) to a reaction explicitly stated to be zero order.
• Trying to use an integrated rate law to compute a "changing" rate when the defining feature of zero order is precisely that the rate does not change.

ANSWER: B

Concept and Intuition

A zero-order reaction has rate independent of concentration: $\text{rate}=k$, a constant. Integrating $-\dfrac{d[\mathrm{R}]}{dt}=k$ gives a straight-line decay of concentration with time, unlike first-order's exponential decay — so, unusually, a zero-order reaction can reach exactly zero concentration in finite time.

Step-by-Step Solution

1. Integrated zero-order rate law: $[\mathrm{R}]=[\mathrm{R}]_0-kt$.
2. "100% completion" means all reactant is consumed: $[\mathrm{R}]=0$.
3. Substitute: $0=[\mathrm{R}]_0-kt \Rightarrow t=\dfrac{[\mathrm{R}]_0}{k}$.

Common Mistakes

• Confusing this with the zero-order half-life formula $t_{1/2}=\dfrac{[\mathrm{R}]_0}{2k}$, which corresponds to 50% completion, not 100%.

ANSWER: C

Concept and Intuition

For a zero-order reaction, $\text{rate}=k[A]^0=k$. This means the rate does not change as the reaction proceeds or as concentration varies — a defining and easily-testable feature of zero-order kinetics.

Step-by-Step Solution

1. Rate law for zero order: $\text{rate}=k$.
2. Given $k=2.5\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}$ at 300 K.
3. The concentration value ($10^{-1}\,\mathrm{mol\,L^{-1}}$) is a distractor — it does not affect the rate for a zero-order reaction.
4. Rate = $2.5\times10^{-4}\,\mathrm{mol\,L^{-1}\,s^{-1}}$.

Common Mistakes

• Multiplying $k$ by the given concentration as if the reaction were first order.

ANSWER: B

Concept and Intuition

For a zero-order reaction $A \to P$, the rate is independent of concentration: $\text{rate}=k$ (constant). Integrating $-\dfrac{d[A]}{dt}=k$ gives the linear law $[A]_t = [A]_0 - kt$ — concentration drops at a constant rate over time, unlike first-order kinetics where it decays exponentially.

Step-by-Step Solution

1. Zero-order integrated rate law: $[A]_t = [A]_0 - kt$.
2. Given: $[A]_0 = 0.1$ mol L$^{-1}$, $k = 1\times10^{-3}$ mol L$^{-1}$ s$^{-1}$, $t = 10$ s.
3. $kt = 1\times10^{-3} \times 10 = 0.01$ mol L$^{-1}$.
4. $[A]_{10} = 0.1 - 0.01 = 0.09$ mol L$^{-1}$.

Common Mistakes

• Applying the first-order exponential decay formula instead of the linear zero-order law.
• Unit slip: forgetting $k$ already has units of concentration/time for a zero-order reaction, so $kt$ directly gives a concentration.

ANSWER: A

Concept and Intuition

For $2NH_3\to N_2+3H_2$, the reaction rate is defined as $\text{Rate}=-\dfrac{1}{2}\dfrac{d[NH_3]}{dt}=\dfrac{d[N_2]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}$. For a zero-order reaction this common rate equals the rate constant $k$ itself (units mol L$^{-1}$s$^{-1}$ match).

Step-by-Step Solution

1. Zero order $\Rightarrow$ Rate $=k=y\times10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$.
2. Rate $=\dfrac13\dfrac{d[H_2]}{dt}\Rightarrow \dfrac{d[H_2]}{dt}=3\times\text{Rate}=3k$.
3. $=3y\times10^{-4}\,\text{mol L}^{-1}\text{s}^{-1}$.

Common Mistakes

• Assuming the rate of formation of H$_2$ equals $k$ directly, ignoring its stoichiometric coefficient of 3.
• Confusing $k$ (given directly with rate units, since it's zero order) with a rate constant that would need multiplying by a concentration term.

ANSWER: C

Concept and Intuition

In a zero-order reaction, the rate is independent of concentration -- the concentration falls linearly with time, unlike first/second order reactions where it falls exponentially or hyperbolically. The integrated rate law is simply $[A]_t = [A]_0 - kt$, a straight line of slope $-k$.

Step-by-Step Solution

1. Zero-order integrated rate law: $[A]_t = [A]_0 - kt$.
2. Given $[A]_0 = 0.10\ M$, $[A]_t = 0.075\ M$, $k = 0.0030\ mol\,L^{-1}\,s^{-1}$.
3. Rearranging: $t = \dfrac{[A]_0 - [A]_t}{k}$.
4. Substitute: $t = \dfrac{0.10 - 0.075}{0.0030} = \dfrac{0.025}{0.0030} = 8.33\ s$.
5. This matches option (C).

Common Mistakes

• Applying the first-order formula ($t = \frac{1}{k}\ln\frac{[A]_0}{[A]}$) by mistake -- the units of $k$ ($mol\,L^{-1}\,s^{-1}$, not $s^{-1}$) are the tell-tale sign that this is a zero-order reaction.

ANSWER: C

Concept and Intuition

For a zero-order reaction, rate $=k$ is constant (independent of concentration), so $[A]_t = [A]_0 - kt$ — a straight-line decay. The rate of formation of product is scaled by the stoichiometric coefficient: since 3 mol of B appear for every 1 mol of A consumed, $\dfrac{d[B]}{dt} = 3\times\left(-\dfrac{d[A]}{dt}\right) = 3k$.

Step-by-Step Solution

1. From the data at $t=0$ ($[A]=0.1$) and $t=15$ s ($[A]=0.05$): $k = \dfrac{0.1-0.05}{15} \approx 0.0033\ \text{mol L}^{-1}\text{s}^{-1}$.
2. Amount of A reacted by $t=20$ s: $x = k\times20 \approx 0.0033\times20 = 0.066\ \text{mol/L}$ (this is the "$x$" in $[A]=0.1-x$).
3. Since $A\to 3B$, moles of B formed $= 3x \approx 3\times0.066 = 0.198\ \text{mol/L} = 1.98\times10^{-1}$ mol/L.

Common Mistakes

• Forgetting the factor of 3 from stoichiometry and reporting just $x$ (the amount of A consumed) instead of $[B]$.
• Applying a first-order (exponential) formula instead of the zero-order linear law.

ANSWER: C

Concept and Intuition

For a zero order reaction, the rate is independent of reactant concentration, so the integrated rate law is linear in time: $[P] = [P]_0 - kt$, where $k$ is the (constant) rate. This means equal time intervals always remove the same amount (not the same fraction) of reactant — unlike first order kinetics where equal time intervals remove the same fraction.

Step-by-Step Solution

1. Use the given data (0.1 M → 0.05 M in 10 s) to find $k$: $k = \dfrac{[P]_0 - [P]}{t} = \dfrac{0.1-0.05}{10} = \dfrac{0.05}{10} = 0.005\ \mathrm{mol\,L^{-1}\,s^{-1}}$.
2. Apply the zero-order integrated law at $t = 15\,\mathrm{s}$ (measuring from the same $t=0$, $[P]_0=0.1$ M): $[P] = [P]_0 - kt = 0.1 - (0.005)(15)$.
3. Compute: $(0.005)(15) = 0.075$, so $[P] = 0.1 - 0.075 = 0.025\ \mathrm{M}$.

Common Mistakes

• Treating the reaction as first order and using a half-life/exponential decay formula instead of the linear zero-order law.
• Mis-starting the 15 s clock from the 10 s mark instead of from $t=0$ (the problem says "concentration of P after 15 seconds time", i.e. from the start).

ANSWER: B

Concept and Intuition

The order of reaction with respect to a species tells you how sensitively the rate depends on that species' concentration: rate $\propto [A]^n$. If increasing $[A]$ has no effect on the rate, the exponent $n$ must be zero, because any nonzero power of 2 (the doubling factor) would change the rate. Zero-order behaviour typically arises when the rate-determining step doesn't actually involve free $A$ in solution — e.g., a heterogeneous catalytic reaction where the catalyst surface is already saturated with A, so adding more A in solution can't speed anything up.

Step-by-Step Solution

1. Write the general rate law: rate $= k[A]^n$.
2. Let the initial concentration be $[A]$, so initial rate $r_1 = k[A]^n$.
3. After doubling, new concentration is $2[A]$, so new rate $r_2 = k(2[A])^n = 2^n \cdot k[A]^n = 2^n r_1$.
4. Given $r_2 = r_1$ (rate unchanged): $2^n r_1 = r_1 \Rightarrow 2^n = 1 \Rightarrow n = 0$.
5. So the reaction is zero order with respect to A.

Common Mistakes

• Confusing the phrase 'rate unchanged' with 'rate doubled' (which would instead indicate first order, $n=1$).
• Assuming order must be a whole positive number greater than zero by default.

ANSWER: D

Concept and Intuition

For a zero-order reaction the rate is constant ($\text{rate}=k$, independent of concentration), so the integrated rate law is linear in time:

Half-life is the time at which $[A]_t = \dfrac{[A]_0}{2}$.

Step-by-Step Solution

1. Set $[A]_t = \dfrac{[A]_0}{2}$ in the integrated law: $\dfrac{[A]_0}{2} = [A]_0 - k t_{1/2}$.
2. Solve: $k t_{1/2} = [A]_0 - \dfrac{[A]_0}{2} = \dfrac{[A]_0}{2}$, so $t_{1/2} = \dfrac{[A]_0}{2k}$.
3. This is of the form $y = mx$ with $y=t_{1/2}$, $x=[A]_0$, and slope $m = \dfrac{1}{2k}$ — a straight line through the origin.

Common Mistakes

• Confusing this with the first-order half-life ($t_{1/2}=\dfrac{0.693}{k}$, which is independent of $[A]_0$ — a horizontal line).
• Mixing up slope $=2k$ vs slope $=\dfrac{1}{2k}$.

ANSWER: D