Decomposition of A is a first order reaction at T(K) and is given by **A(g) → B(g) + C(g).**

In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min$^{-1}$) of the reaction? (log 2 = 0.3)

Question

Decomposition of A is a first order reaction at T(K) and is given by **A(g) → B(g) + C(g).**

In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min$^{-1}$) of the reaction? (log 2 = 0.3)

Accepted Answer

D. 1. **Set up the pressure relationships.** Given A decomposes to B and C in equal proportions, if $x$ bar of A decomposes,
$$
P_{total} = (1-x)+x+x = 1+x
$$

2. **Solve for x using the given total pressure.** Given after 100 min, $P_{total}=1.5$ bar,
$$
1+x=1.5 \quad \Rightarrow \quad x=0.5
$$

3. **Find the remaining pressure of A.** Given
$$
P_A = 1-x = 0.5\ \mathrm{bar}
$$

4. **Apply the first-order rate law using pressure as a proxy for concentration.** Given
$$
k = \frac{2.303}{t}\log\left(\frac{P_{A,0}}{P_{A,t}}ight) = \frac{2.303}{100}\log\left(\frac{1}{0.5}ight) = \frac{2.303}{100}	imes 0.3 = 6.9	imes 10^{-3}\ \mathrm{min}^{-1}
$$

**Hence, the answer is option D.**

Decomposition of A is a first order reaction at T(K) and is given by A(g) → B(g) + C(g). In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min−1^{-1}−1) of the reaction? (log 2 = 0.3)

Decomposition of A is a first order reaction at T(K) and is given by A(g) → B(g) + C(g).

In a closed 1 L vessel, 1 bar A(g) is allowed to decompose at T(K). After 100 minutes, the total pressure was 1.5 bar. What is the rate constant (in min $^{-1}$ ) of the reaction? (log 2 = 0.3)