Write the slope value obtained in the plot of ln[R] vs. time for a first order reaction
Question
Write the slope value obtained in the plot of ln[R] vs. time for a first order reaction
Solution
In a first order reaction, the plot of ln[R] (natural logarithm of the concentration of the reactant) versus time gives a straight line with a negative slope. The value of this slope is equal to the negative of the rate constant (k) of the reaction.
Here are the steps to obtain the slope:
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Plot the natural logarithm of the concentration of the reactant (ln[R]) on the y-axis and time on the x-axis.
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Draw the best fit straight line through the data points.
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Determine the slope of this line. The slope of a line is given by the change in y divided by the change in x (Δy/Δx). In this case, it would be the change in ln[R] divided by the change in time.
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The slope obtained will be a negative number, which is the negative of the rate constant (k) for the reaction.
So, the slope value obtained in the plot of ln[R] vs. time for a first order reaction is -k.
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Rate of Reaction: Rate = - Δ[Reactant]/Δt = Δ[Product]/Δt Rate Law: Rate = k[A]^x[B]^y First Order Reaction: ln[A]t = -kt + ln[A]0 Second Order Reaction: 1/[A]t = kt + 1/[A]0 Zero Order Reaction: [A]t = -kt + [A]0 Half-Life (t1/2) for First Order Reaction: t1/2 = 0.693/k Half-Life (t1/2) for Second Order Reaction: t1/2 = 1/(k[A]0) Half-Life (t1/2) for Zero Order Reaction: t1/2 = [A]0/2k Arrhenius Equation: k = Ae^(-Ea/RT) where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. Integrated Rate Law for First Order Reaction: ln([A]0/[A]t) = kt Integrated Rate Law for Second Order Reaction: 1/[A]t - 1/[A]0 = kt Integrated Rate Law for Zero Order Reaction: [A]0 - [A]t = kt
A rate graph has a steep gradient initially followed by a decreased gradient that eventually flattens out. Which point on the graph indicates the time that the reaction stopped?
Analyse the given graph, drawn between concentration of reactant vs. time.(a) Predict the order of reaction.
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