For the reaction, H2 + I2 - 2HI, the differential rate law is
(a) -d[H2]/dt = -d[I2]/dt = 2d[HI]/dt

Q: Why do we use a negative sign for reactants?

Reactants are consumed during the reaction, so their concentration decreases over time. The negative sign ensures that the overall rate of reaction remains a positive quantity.

Q: Why do we divide by the stoichiometric coefficient?

This normalizes the rate of reaction so that it is the same regardless of which reactant or product you are measuring. For example, if 2 moles of HI are produced for every 1 mole of H2 consumed, HI appears twice as fast as H2 disappears.

Q: Can we multiply the rate equation by any number?

Yes, the rate equation is an equality. Multiplying all terms by the same non-zero constant maintains the equality, which is often needed to match multiple-choice options.

Q: What does $d[H_2]/dt$ represent?

It represents the instantaneous rate of change of the molar concentration of hydrogen gas with respect to time.

Q: Is the rate of reaction always positive?

Yes, the rate of a reaction is defined as a positive value, which is why we add negative signs to the rates of change of reactants.

Question

For the reaction, $H_2 + I_2 \longrightarrow 2HI$, the differential rate law is

Options:

(A) $-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = 2\frac{d[HI]}{dt}$

(B) $-2\frac{d[H_2]}{dt} = -2\frac{d[I_2]}{dt} = \frac{d[HI]}{dt}$

(C) $-\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{d[HI]}{dt}$

(D) $-\frac{d[H_2]}{2dt} = -\frac{d[I_2]}{2dt} = \frac{d[HI]}{dt}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a closed vessel where hydrogen gas ($H_2$) and iodine gas ($I_2$) are reacting to form hydrogen iodide ($2HI$). As the reaction proceeds, the reactants are being consumed, and the product is being formed. The speed at which this happens is what we call the **rate of reaction**.

To express this mathematically, we use the differential rate law. The rate of a reaction must be a positive value and should be the same regardless of which species we are monitoring.

### The Master Equation
For a general reaction $aA + bB \longrightarrow cC$, the rate of reaction is given by:
$$ 	ext{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} $$

Notice the negative signs for the reactants. Since their concentration is decreasing, the derivative is negative. The negative sign ensures the overall rate is positive. We also divide by the stoichiometric coefficients to normalize the rate.

### Final Calculation
Let's apply this to our specific reaction: $H_2 + I_2 \longrightarrow 2HI$.
The stoichiometric coefficients are 1 for $H_2$, 1 for $I_2$, and 2 for $HI$.

Substituting these into our master equation, we get:
$$ 	ext{Rate} = -\frac{d[H_2]}{dt} = -\frac{d[I_2]}{dt} = \frac{1}{2}\frac{d[HI]}{dt} $$

Now, we look at our options. None of them match this exact form. But remember, this is an algebraic equation. We can multiply the entire equation by a constant without changing its meaning. Let's multiply everything by 2:
$$ -2\frac{d[H_2]}{dt} = -2\frac{d[I_2]}{dt} = \frac{d[HI]}{dt} $$

**This perfectly matches option (b)!** By simply scaling our equation, we found the correct representation of the differential rate law.

Animated Solution for Chemistry - Chemical Kinetics: For the reaction, $H_{2} + I_{2} ⟶ 2 H I$ , the differential rate law is

Select Answer:

Visualized Solution

$H_{2} + I_{2} ⟶ 2 H I$

$Rate of Reaction$

$Writing the Rate Law$

$Matching the Options$

$Conclusion$

$Physical Significance$

The Sigma Insight: Rate of Chemical Reaction

Analyzing the Setup Imagine a closed vessel where hydrogen gas ( $H_{2}$ ) and iodine gas ( $I_{2}$ ) are reacting to form hydrogen iodide ( $2 H I$ )

The Master Equation

Final Calculation Let's apply this to our specific reaction: $H_{2} + I_{2} ⟶ 2 H I$

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Animated Solution for Chemistry - Chemical Kinetics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For the reaction, H2​+I2​⟶2HI, the differential rate law is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​+I2​⟶2HI

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rate of Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Writing the Rate Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Matching the Options

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Significance

The Sigma Insight: Rate of Chemical Reaction

The Master Equation

Final Calculation Let's apply this to our specific reaction: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H2​+I2​⟶2HI

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