Decomposition of H2O2 follows a first order reaction. In 50 min, the concentration of H2O2 decreases

Q: Why did we use the half-life method instead of the standard first-order equation?

While you can absolutely use the standard equation $k = \frac{2.303}{t} \log\left(\frac{a}{a-x}\right)$, noticing that 0.125 is exactly one-fourth of 0.5 saves you a lot of calculation time. It's a classic JEE shortcut!

Q: Why is the rate of formation of $O_2$ half the rate of disappearance of $H_2O_2$?

Look at the balanced equation: $2H_2O_2 \rightarrow 2H_2O + O_2$. For every 2 moles of $H_2O_2$ that decompose, only 1 mole of $O_2$ is formed. Therefore, $O_2$ is formed at half the speed that $H_2O_2$ disappears.

Q: Can I use the rate constant $k$ directly to find the rate of formation of $O_2$?

No, the rate constant $k$ gives you the rate of disappearance of $H_2O_2$ when multiplied by its concentration (Rate $= k[H_2O_2]$). You must apply the stoichiometric coefficient to find the rate of formation of $O_2$.

Q: What does the tangent on the concentration-time graph represent?

The slope of the tangent at any point on a concentration-time graph gives the instantaneous rate of reaction at that specific moment. In our case, it represents the rate of disappearance of $H_2O_2$ at $[H_2O_2] = 0.05$ M.

Q: Why didn't we convert minutes to seconds?

Always look at the units in the options! All the options are given in terms of $\text{min}^{-1}$, so converting to seconds would just add unnecessary steps and potential for errors.

Question

Decomposition of $H_2O_2$ follows a first order reaction. In $50 	ext{ min}$, the concentration of $H_2O_2$ decreases from $0.5$ to $0.125 	ext{ M}$ in one such decomposition. When the concentration of $H_2O_2$ reaches $0.05 	ext{ M}$, the rate of formation of $O_2$ will be

Options:

(A) $6.93 	imes 10^{-4} 	ext{ mol min}^{-1}$

(B) $2.66 	ext{ L min}^{-1} 	ext{ at STP}$

(C) $1.34 	imes 10^{-2} 	ext{ mol min}^{-1}$

(D) $6.93 	imes 10^{-2} 	ext{ mol min}^{-1}$

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Decay

Imagine you are observing the decomposition of hydrogen peroxide ($H_2O_2$) in a beaker. As time passes, its concentration drops exponentially, which is a classic signature of first-order kinetics. The balanced chemical equation for this process is:

$$2H_2O_2 ightarrow 2H_2O + O_2$$

This equation is our roadmap. It tells us exactly how the disappearance of our reactant is linked to the appearance of our products.

### The Half-Life Shortcut

The problem states that the concentration of $H_2O_2$ drops from $0.5 	ext{ M}$ to $0.125 	ext{ M}$ in $50 	ext{ minutes}$. While you could plug these numbers into the standard first-order integrated rate law, there is a much faster, more elegant way to see this.

Notice the numbers: $0.5$ halves to $0.25$, and $0.25$ halves again to $0.125$. This means exactly **two half-lives** have passed!

$$0.5 \xrightarrow{t_{1/2}} 0.25 \xrightarrow{t_{1/2}} 0.125$$

Since two half-lives take $50 	ext{ minutes}$, a single half-life ($t_{1/2}$) must be $25 	ext{ minutes}$. Now, we can easily find the rate constant, $k$, using the half-life formula for first-order reactions:

$$k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{25} 	ext{ min}^{-1}$$

We will keep it in this fraction form for now to avoid messy intermediate calculations.

### The Stoichiometric Link

Now, let's look at the stoichiometry of the reaction. Two moles of hydrogen peroxide produce one mole of oxygen. Therefore, the rate of formation of oxygen is exactly half the rate of disappearance of hydrogen peroxide.

$$	ext{Rate of reaction} = -\frac{1}{2} \frac{d[H_2O_2]}{dt} = \frac{d[O_2]}{dt}$$

Since it is a first-order reaction, the rate of disappearance of $H_2O_2$ is simply $k$ times its concentration:

$$-\frac{d[H_2O_2]}{dt} = k[H_2O_2]$$

Substituting this into our rate equation gives us the master equation for the rate of formation of oxygen:

$$\frac{d[O_2]}{dt} = \frac{1}{2} k [H_2O_2]$$

### The Final Calculation

The question asks for the rate at the specific instant when the concentration of hydrogen peroxide reaches $0.05 	ext{ M}$. On a concentration-time graph, this corresponds to finding the slope of the tangent line at this exact point.

Let's substitute the values we have into our master equation:

$$\frac{d[O_2]}{dt} = \frac{1}{2} 	imes \left( \frac{0.693}{25} ight) 	imes 0.05$$

Now for the final calculation. Two times twenty-five in the denominator gives fifty. Multiplying $0.693$ by $0.05$ and dividing by $50$ gives us:

$$\frac{d[O_2]}{dt} = \frac{0.03465}{50} = 6.93 	imes 10^{-4} 	ext{ mol min}^{-1}$$

And there we have it! The rate of formation of oxygen at that exact moment is **$6.93 	imes 10^{-4} 	ext{ mol min}^{-1}$**.

Select Answer:

Visualized Solution

$[H_{2} O_{2}]$ vs $t$

$t_{1/2}$

$k$

$\frac{d [ O _{2} ]}{d t}$

$[H_{2} O_{2}] = 0.05 M$

$\frac{d [ O _{2} ]}{d t}$

$\frac{d [ O _{2} ]}{d t}$

$\frac{d [ H _{2} O ]}{d t}$

The Sigma Insight: Rate of Chemical Reaction

The Setup

The Half-Life Shortcut

The Stoichiometric Link

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [H2​O2​] vs .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t1/2​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtd[O2​]​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [H2​O2​]=0.05 M

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtd[O2​]​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtd[O2​]​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } dtd[H2​O]​

The Sigma Insight: Rate of Chemical Reaction

The Setup

The Half-Life Shortcut

The Stoichiometric Link

The Final Calculation

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$[H_{2} O_{2}]$ vs $t$

$t_{1/2}$

$k$

$\frac{d [ O _{2} ]}{d t}$

$[H_{2} O_{2}] = 0.05 M$

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For the reaction $2 H_{2} (g) + 2 NO (g) \to N_{2} (g) + 2 H_{2} O (g)$ the observed rate expression is, $rate = K_{f} [NO]^{2} [H_{2}]$ . The rate expression for the reverse reaction is