Q.Rate law cannot be determined from balanced chemical equation if _______. (Two or more than two options may be correct.)

The key idea here is that the rate law of a reaction is an experimentally determined expression — it cannot be reliably predicted from the balanced chemical equation unless the reaction is elementary (occurs in a single step). For multi-step (complex) reactions, the rate law depends on the slowest step, not the overall stoichiometry.

Reasoning:

1. For an elementary reaction, the rate law can be written directly from the balanced equation (using stoichiometric coefficients as exponents). So option (ii) is not a case where rate law cannot be determined — it can be.
2. If the reaction is a sequence of elementary reactions (complex reaction), the rate law is governed by the rate-determining step, not the overall equation. Hence (iii) is correct.
3. If a reverse reaction is significant, the net rate involves both forward and backward rates, and the simple form from the balanced equation fails — so (i) is correct. …

The balanced chemical equation tells you the stoichiometric coefficients — the numbers that balance atoms and charge. But the rate law tells you how the rate depends on concentration, and that depends on the mechanism: the actual sequence of molecular collisions or steps. Unless the reaction happens in one single step (an elementary reaction), the coefficients in the balanced equation have no direct relation to the exponents in the rate law.

Let’s walk through each option carefully.

1. Option (i): reverse reaction is involved.

   If the reverse reaction is significant, the net rate is the difference between forward and reverse rates. The forward rate might follow one law, the reverse another, and the overall observed rate law can look very different from what the balanced equation suggests. For example, in a reversible reaction $A \rightleftharpoons B$, the net rate near equilibrium depends on both $[A]$ and $[B]$, not just on the stoichiometry of $A \rightarrow B$. So yes — when a reverse reaction is involved, you cannot deduce the rate law from the balanced equation alone.

2. Option (ii): it is an elementary reaction.

   This is the exception. For an elementary reaction, the reaction occurs exactly as written in one step. In that case, the rate law is directly given by the stoichiometric coefficients. For example, if $2A + B \rightarrow C$ is elementary, then rate $= k[A]^2[B]$. So here the rate law can be determined from the balanced equation. This option is not correct for the blank.

3. Option (iii): it is a sequence of elementary reactions.

   Most reactions are multi-step. The overall balanced equation is just the net result. The slowest step (rate-determining step) controls the rate, and its molecularity may involve species that appear nowhere in the overall equation (like intermediates). For instance, the reaction $2NO + O_2 \rightarrow 2NO_2$ actually proceeds via $NO + NO \rightleftharpoons N_2O_2$ (fast) then $N_2O_2 + O_2 \rightarrow 2NO_2$ (slow), giving rate $= k[NO]^2[O_2]$ — which happens to match the stoichiometry here, but that’s a coincidence. In general, you cannot predict the exponents from the balanced equation for a multi-step mechanism. So this is correct.

4. Option (iv): any of the reactants is in excess. …

Method: Conceptual Elimination Using the Definition of Rate Law

The rate law of a reaction is an experimentally determined equation that shows how the rate depends on the concentration of reactants. It cannot be predicted from the balanced chemical equation unless the reaction is elementary (occurs in a single step).

Step-by-step reasoning:

1. Recall the key principle

   • For an elementary reaction, the rate law can be written directly from the balanced equation (using stoichiometric coefficients as exponents).
   • For complex reactions (sequences of elementary steps), the rate law depends on the slowest step — not on the overall balanced equation.

2. Analyze each option

   • (i) Reverse reaction is involved

     If the reverse reaction is significant, the net rate is affected by both forward and backward rates. The simple form of the rate law (from the forward reaction alone) does not hold.

     → Correct (rate law cannot be determined from the balanced equation alone).

   • (ii) It is an elementary reaction

     For an elementary reaction, the rate law can be determined directly from the balanced equation.

     → Incorrect (this is the exception where it is possible).

   • (iii) It is a sequence of elementary reactions

     The overall balanced equation gives no information about the mechanism. The rate law depends on the slowest step, not the overall stoichiometry.

     → Correct.

   • (iv) Any of the reactants is in excess …

Common Mistakes Students Make on This Question

Mistake 1: Thinking the rate law can be determined from a balanced equation for elementary reactions

Why it's wrong:

The question asks when the rate law cannot be determined. For an elementary reaction, the rate law can be determined directly from the balanced equation (using stoichiometric coefficients as exponents). So option (ii) is not a case where the rate law cannot be determined.

How to avoid:

Remember the key rule:

• Elementary reaction → Rate law = $k[\text{reactant}_1]^{\text{coefficient}_1}[\text{reactant}_2]^{\text{coefficient}_2}...$
• Complex reaction (sequence of elementary steps) → Rate law depends on the slowest step, not the overall balanced equation.

Mistake 2: Confusing "reverse reaction" with "reversible reaction" in rate law context

Why it's wrong:

If a reverse reaction is involved, the net rate is the difference between forward and reverse rates. The balanced equation alone cannot give you the rate law because the reverse step also contributes. So option (i) is correct.

How to avoid:

When you see "reverse reaction," think:

• Forward rate = $k_f[\text{A}]^a[\text{B}]^b$
• Reverse rate = $k_r[\text{C}]^c[\text{D}]^d$
• Net rate = forward − reverse The balanced equation doesn't tell you $k_f$, $k_r$, or the exponents — so rate law cannot be determined.

Mistake 3: Thinking "excess reactant" means rate law can still be determined

Why it's wrong:

When a reactant is in excess, its concentration remains nearly constant during the reaction. This simplifies the rate law (pseudo-order), but the actual rate law still cannot be deduced from the balanced equation alone — you need experimental data. So option (iv) is correct.

How to avoid:

Remember:

• Excess reactant → pseudo-order kinetics (e.g., $R = k'[\text{A}]^m$ where $k' = k[\text{B}]_0^n$)
• But the balanced equation doesn't tell you $m$ or $n$ — only experiments can.

Mistake 4: Missing that "sequence of elementary reactions" is a correct answer

Why it's wrong:

A complex reaction (sequence of elementary steps) has a rate law determined by the slowest step, not the overall balanced equation. So option (iii) is correct.

How to avoid: …