The graph shows a linear relationship between log |R₀|
and time. This suggests a first-order reaction.
In a first-order reaction, the concentration of the reactant decreases exponentially over time. This means:
\[ \ln [R] = \ln [R_0] - kt \]
Taking the logarithm (base 10) of both sides also gives a linear equation, which justifies that a plot of \(\log [R]\) versus time is a straight line.
For a first-order reaction, the slope of the \(\log [R]\) vs time graph is equal to \(-k\), where \(k\) is the rate constant.
So, the slope \(m\) of the graph is:
\[ \text{slope} = -k \]
Time (Hours) | [A] (M) |
---|---|
0 | 0.40 |
1 | 0.20 |
2 | 0.10 |
3 | 0.05 |
The decomposition of a compound A follows first-order kinetics. The concentration of A at time t = 0 is 1.0 mol L-1. After 60 minutes, it reduces to 0.25 mol L-1. What is the initial rate of the reaction at t = 0? (Take ln 2 = 0.693)