Decomposition of A is a first order reaction at T(K) and is given by \( A(g) \rightarrow B(g) + C(g) \).
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 ? (\( \log 2 = 0.3 \))
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For a reaction \( A \rightarrow nB \), the total pressure relation is \( P_t = P_o + (n-1)x \). This helps you find the partial pressure of reactant quickly.
The given problem involves the decomposition of a gas \( A \) in a first-order reaction, and we are required to calculate the rate constant (\( k \)) of this reaction. The reaction is as follows:
\(A(g) \rightarrow B(g) + C(g)\)
Here's how we can solve the problem step-by-step:
Initially, in a closed 1 L vessel, the pressure of \( A \) is 1 bar.
After 100 minutes, the total pressure in the vessel is 1.5 bar. Let's say \( x \) bar of gas \( A \) has decomposed. Therefore, as 1 mole of \( A \) produces 1 mole of \( B \) and 1 mole of \( C \), the increase in total pressure due to the formation of \( B \) and \( C \) is \( 2x \).
The initial pressure (1 bar) minus the change in pressure (\( x \)) gives the remaining pressure of \( A \). Therefore, the remaining pressure of \( A \) is \( 1 - x \).
The total pressure at 100 minutes being 1.5 bar means:
Solving for \( x \), we find:
Thus, the change in pressure (or the amount of \( A \) decomposed) is 0.5 bar.
The remaining pressure of \( A \), therefore, is:
For a first-order reaction, the rate constant formula is given by:
Substitute the values where \( t = 100 \, \text{minutes} \), \( P_0 = 1 \, \text{bar} \), and \( P_t = 0.5 \, \text{bar} \):
Therefore, the rate constant \( k \) of the reaction is approximately \(6.9 \times 10^{-3} \, \text{min}^{-1}\), which matches the correct option given.
