
In a first-order reaction, the instantaneous rate of reaction is given by the equation: \[ \text{Rate} = k [A] \] The slope of the graph at point C represents the rate constant \( k \), and thus, the instantaneous rate is directly related to the slope. The slope \( m \) of the graph represents the rate.
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?
In the given circuit, if the potential at point B is 24 V, the potential at point A is:
