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Gaseous First Order Reactions

Calculating Rate Constants using Partial and Total Pressures.

By chemca Team • Updated Jan 2026

For gaseous reactions occurring at constant temperature, the concentration is directly proportional to the partial pressure ($P \propto C$ since $PV=nRT \rightarrow P = [C]RT$). Therefore, we can express rate laws in terms of atmospheric pressure instead of molarity.

1. Typical Decomposition ($A \rightarrow B + C$)

Derivation

Consider a first-order gas phase reaction where reactant A decomposes into two gaseous products B and C.

$$ A(g) \rightarrow B(g) + C(g) $$

Time	Pressure of A ($P_A$)	Pressure of B ($P_B$)	Pressure of C ($P_C$)	Total Pressure ($P_t$)
$t = 0$	$P_0$ (Initial)	0	0	$P_0$
$t = t$	$P_0 - x$	$x$	$x$	$(P_0 - x) + x + x = P_0 + x$

Step 1: Calculate $x$ in terms of $P_t$ and $P_0$.

$$ P_t = P_0 + x \implies x = P_t - P_0 $$

Step 2: Find pressure of A at time t ($P_A$).

$$ P_A = P_0 - x = P_0 - (P_t - P_0) = 2P_0 - P_t $$

Step 3: Substitute into Integrated Rate Equation.

$$ k = \frac{2.303}{t} \log \left( \frac{P_{initial}}{P_{final}} \right) $$ $$ k = \frac{2.303}{t} \log \left( \frac{P_0}{2P_0 - P_t} \right) $$

2. General Stoichiometry

General Formula

Consider the reaction: $A(g) \rightarrow nB(g) + mC(g)$

Let $x$ be the decrease in pressure of A.
Total Pressure $P_t = (P_0 - x) + nx + mx = P_0 + x(n+m-1)$.

Logic Tip: Do not memorize complex formulas for every stoichiometry. Always follow the table method:

1. Write the reaction.
2. Define pressures at $t=0$ and $t=t$ using '$x$'.
3. Sum them to get $P_t$.
4. Solve for $x$.
5. Substitute $x$ back into $P_0 - x$ (Pressure of Reactant).

3. Common Examples

A. Decomposition of $N_2O_5$

$$ 2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g) $$

At time t: $P_{N_2O_5} \propto (P_0 - 2x)$, $P_{N_2O_4} \propto 2x$, $P_{O_2} \propto x$.

$$ P_t = (P_0 - 2x) + 2x + x = P_0 + x \implies x = P_t - P_0 $$ $$ P_{reactant} = P_0 - 2x = P_0 - 2(P_t - P_0) = 3P_0 - 2P_t $$ $$ k = \frac{2.303}{t} \log \left( \frac{P_0}{3P_0 - 2P_t} \right) $$

B. Decomposition of Azoisopropane

$$ (CH_3)_2CHN=NCH(CH_3)_2(g) \rightarrow N_2(g) + C_6H_{14}(g) $$

Stoichiometry: $1 \rightarrow 1 + 1$. This matches the standard case derived in Section 1.

$$ k = \frac{2.303}{t} \log \left( \frac{P_0}{2P_0 - P_t} \right) $$

4. Pressure at Infinite Time ($P_\infty$)

At $t = \infty$, the reaction is complete. Reactant pressure becomes 0.

For $A \rightarrow B + C$:

$$ P_\infty = 0 + P_0 + P_0 = 2P_0 \implies P_0 = \frac{P_\infty}{2} $$

You can substitute $P_0$ in terms of $P_\infty$ if initial pressure is not given but final pressure is.

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