Q.How many mL of 0.1 M HCl are required to react completely with 1 g mixture of $Na_2CO_3$ and $NaHCO_3$ containing equimolar amounts of both?

Concept and Intuition

For a reaction $2A \to$ products, the rate of reaction is not simply the rate of disappearance of A — it must be divided by A's stoichiometric coefficient so that the rate is the same regardless of which species you track: $\text{Rate} = -\dfrac{1}{2}\dfrac{d[A]}{dt}$.

Step-by-Step Solution

1. Change in concentration of A: $\Delta[A] = 0.4-0.5 = -0.1$ M over $\Delta t = 10$ min.
2. Rate of disappearance of A $= \dfrac{0.1}{10} = 0.01$ mol L⁻¹ min⁻¹. …

Concept and Intuition

For a reaction with known stoichiometry, measuring how much of ONE product forms lets you find how much reactant disappeared, using mole ratios from the balanced equation. Average rate over a finite time interval is simply $\Delta(\text{concentration})/\Delta t$, referenced to whichever species' stoichiometric coefficient is being tracked (here, N$_2$O$_5$, coefficient 1).

Step-by-Step Solution

1. Initial moles of $N_2O_5$: $50\,\text{mL} \times 2\,M = 0.1$ mol; initial concentration $=2$ M.
2. Moles of $O_2$ formed at STP: $\dfrac{0.28\,L}{22.4\,L/mol} = 0.0125$ mol.
3. From $N_2O_5 \rightarrow N_2O_4 + \tfrac12 O_2$: every 1 mol $O_2$ formed corresponds to 2 mol $N_2O_5$ consumed. So $N_2O_5$ consumed $= 2\times0.0125 = 0.025$ mol.
4. Unreacted $N_2O_5$ $= 0.1-0.025 = 0.075$ mol; in 50 mL this is $X = 0.075/0.050 = 1.5$ M. …

Concept and Intuition

For a general reaction, the single unambiguous "rate of reaction" is defined by dividing each species' rate of change by its own stoichiometric coefficient: $\text{Rate} = -\dfrac{1}{a}\dfrac{d[A]}{dt} = -\dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt}$. This ensures the reported rate is the same number regardless of which species you measure, since a species with a larger coefficient appears/disappears proportionally faster.

Step-by-Step Solution

1. For $A + 2B \rightarrow 3C + 2D$, the rate of reaction is $\text{Rate} = -\dfrac{d[A]}{dt} = -\dfrac{1}{2}\dfrac{d[B]}{dt} = \dfrac{1}{3}\dfrac{d[C]}{dt} = \dfrac{1}{2}\dfrac{d[D]}{dt}$.
2. So Rate of appearance of C $= \dfrac{d[C]}{dt} = 3 \times \text{Rate of reaction}$.
3. Therefore, $\dfrac{\text{Rate of reaction}}{\text{Rate of appearance of C}} = \dfrac{\text{Rate}}{3\,\text{Rate}} = \dfrac{1}{3}$. …

Concept and Intuition

Metal bicarbonates decompose on heating to the carbonate, releasing water and carbon dioxide: $2MHCO_3 \rightarrow M_2CO_3 + H_2O + CO_2$. That released CO2, when passed into a metal hydroxide solution in a 1:1 ratio (as opposed to a 1:2 ratio which would give the carbonate), reforms the bicarbonate: $CO_2+MOH\rightarrow MHCO_3$. This is analogous to the well-known $CO_2+NaOH\rightarrow NaHCO_3$ reaction (in excess CO2).

Step-by-Step Solution

1. From decomposition, moles $CO_2$ produced $=\tfrac12\times$ (moles $MHCO_3$ initially).
2. From the CO2 + MOH reaction (1:1 stoichiometry), moles $CO_2$ = moles $MOH$ reacted completely $= 0.2$ mol.
3. So moles of original $MHCO_3 = 2\times0.2=0.4$ mol. …

Concept and Intuition

For a reaction aA + bB → cC + dD, the rate of reaction is defined so that it's independent of which species you track: Rate = −(1/a)Δ[A]/Δt = −(1/b)Δ[B]/Δt = (1/c)Δ[C]/Δt = ... Each species' own rate of change must be divided by its stoichiometric coefficient.

Step-by-Step Solution

1. Reaction: 5Br⁻ + BrO3⁻ + 6H⁺ → 3Br2 + 3H2O, so Br⁻ has stoichiometric coefficient 5.
2. Given: −Δ[Br⁻]/Δt = x mol L⁻¹ min⁻¹.
3. Overall rate = −(1/5)Δ[Br⁻]/Δt = (1/5)·x = x/5.

Common Mistakes …

Concept and Intuition

For a reaction $2A \rightleftharpoons 4B + C$, the overall rate of reaction is the same whichever species you track, once you divide by its stoichiometric coefficient: $\text{Rate} = -\dfrac{1}{2}\dfrac{d[A]}{dt} = \dfrac{1}{4}\dfrac{d[B]}{dt} = \dfrac{d[C]}{dt}$.

Step-by-Step Solution

1. Rate of increase of B: $\dfrac{\Delta[B]}{\Delta t} = \dfrac{5\times10^{-3}}{10} = 5\times10^{-4}$ mol L⁻¹ s⁻¹.
2. Overall rate of reaction $= \dfrac{1}{4}\times \dfrac{d[B]}{dt} = \dfrac{1}{4}\times5\times10^{-4} = 1.25\times10^{-4}$ mol L⁻¹ s⁻¹. …

Concept and Intuition

For a balanced reaction, the rates of consumption/formation of different species are not numerically equal — they're tied together through the stoichiometric coefficients, because for every 3 molecules of X consumed, exactly 2 molecules of Y are formed (and 1 of Z). Dividing each species' rate of change by its own coefficient gives one common "rate of reaction" that all species share.

Step-by-Step Solution

1. Write the rate of reaction using stoichiometric coefficients: $r=-\dfrac{1}{3}\dfrac{d[X]}{dt}=\dfrac{1}{2}\dfrac{d[Y]}{dt}=\dfrac{d[Z]}{dt}$.
2. Given $-\dfrac{d[X]}{dt}=7.2\times10^{-3}$ mol L⁻¹s⁻¹, compute $r=\dfrac{7.2\times10^{-3}}{3}=2.4\times10^{-3}$ mol L⁻¹s⁻¹. …

Concept and Intuition

Stoichiometric rate relations ($-\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{b}\frac{d[B]}{dt}$) are strictly in terms of moles, not mass. Since the rates here are given in g L$^{-1}$s$^{-1}$, we must first convert to mol L$^{-1}$s$^{-1}$ using molar masses, apply the mole ratio from the balanced equation, then convert the answer back to mass units using the product's molar mass.

Step-by-Step Solution

1. Reaction: $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$.
2. Rate of disappearance of H$_2$ (mass) $= 6\,\text{g L}^{-1}\text{s}^{-1}$. Molar mass of H$_2 = 2\,\text{g/mol}$, so in moles: $6/2 = 3\,\text{mol L}^{-1}\text{s}^{-1}$. …

Concept and Intuition

For a reaction $a A \to b B$, the rates of consumption/production of each species are tied together through the coefficients: $-\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt}$ (the single, unique "rate of reaction"). A species with a larger coefficient changes concentration proportionally faster.

Step-by-Step Solution

1. Reaction: $2\mathrm{N_2O_5(g)} \to 4\mathrm{NO_2(g)} + \mathrm{O_2(g)}$.
2. Rate of reaction $= -\dfrac{1}{2}\dfrac{d[\mathrm{N_2O_5}]}{dt} = \dfrac{1}{4}\dfrac{d[\mathrm{NO_2}]}{dt}$.
3. Given $-\dfrac{d[\mathrm{N_2O_5}]}{dt} = 1.2\times10^{-5}\ \mathrm{mol\,L^{-1}s^{-1}}$, so rate of reaction $= \dfrac{1.2\times10^{-5}}{2} = 0.6\times10^{-5}$.
4. $\dfrac{d[\mathrm{NO_2}]}{dt} = 4 \times (\text{rate of reaction}) = 4 \times 0.6\times10^{-5} = 2.4\times10^{-5}\ \mathrm{mol\,L^{-1}s^{-1}}$. …

Concept and Intuition

In a mechanism made of several elementary steps, the overall reaction rate is controlled by the slowest step because it acts as a bottleneck — no matter how fast the later steps are, the overall reaction cannot proceed faster than the slow step allows. For an elementary step, the rate law can be written directly from its molecularity (unlike for the overall multi-step reaction, where the rate law must be determined experimentally or derived from the mechanism).

Step-by-Step Solution

1. Identify the rate-determining step: since Step 1 is explicitly labelled "slow" and Step 2 is "fast," Step 1 controls the overall rate.
2. Step 1 is $NO_2 + NO_2 \xrightarrow{slow} NO + NO_3$ — this is a bimolecular elementary step involving two molecules of $NO_2$.
3. For an elementary step, the rate law follows directly from the stoichiometry of that step: $rate = k[NO_2][NO_2] = k[NO_2]^2$.
4. The fast second step, $NO_3+CO \to CO_2+NO_2$, consumes the intermediate $NO_3$ as fast as it forms, so it does not appear in (and does not limit) the overall rate expression. …