
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 |
Reactant ‘A’ underwent a decomposition reaction. The concentration of ‘A’ was measured periodically and recorded in the table given below:
Based on the above data, predict the order of the reaction and write the expression for the rate law.
For the reaction \( A + B \to C \), the rate law is found to be \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and \( B \) is halved, by what factor does the rate change?

Simar, Tanvi and Umara were partners in a firm sharing profits and losses in the ratio of 5:6:9. On 31st March, 2024 their Balance Sheet was as follows:

Umara died on 30th June, 2024. The partnership deed provided for the following on the death of a partner: