Chemistry · Chemical Kinetics · NEET
Start from the Arrhenius equation k = A e^(-Ea/RT). Take natural log of both sides: ln k = ln A - (Ea/R)(1/T). Compare this with a straight line y = mx + c, where y = ln k and x = 1/T. So the slope m = -Ea/R and the intercept c = ln A. The slope is always negative because Ea and R are both positive numbers.
The slope = -Ea/R. Rearrange to get Ea = -(slope) x R. Since the slope is a negative number, multiplying by the minus sign makes Ea positive. Use R = 8.314 J K^-1 mol^-1. Example: if slope = -5000 K, then Ea = -(-5000) x 8.314 = 41570 J/mol = 41.5 kJ/mol. This is a very common NEET calculation.
Because ln k = ln A - (Ea/R)(1/T) has the exact form of a straight line y = mx + c. As long as Ea and A do not change with temperature, the equation is linear in the variables ln k and 1/T. That is why NEET asks you to identify the graph that is a straight line with a negative slope (falling from left to right).
As temperature T increases, 1/T decreases (moves left on the x-axis), and k increases so ln k increases (moves up). So higher temperature is on the left-top and lower temperature is on the right-bottom. The line falls as you move right, giving a negative slope. A higher Ea makes the line steeper (more negative slope).
The intercept is ln A, where A is the Arrhenius frequency factor (also called the pre-exponential factor). You read it where the line would meet the y-axis, meaning where 1/T is extended to 0. Some NEET questions give you the intercept value to find A or to solve for the temperature at a given k.
If you plot log k vs 1/T, the slope becomes -Ea/(2.303 R), because ln k = 2.303 log k. So for log-base-10 graphs, Ea = -(slope) x 2.303 x R. Always check whether the axis says ln k (natural log, slope = -Ea/R) or log k (base 10, slope = -Ea/2.303R). This is a classic trap.
Use the two-point form: log(k2/k1) = (Ea / 2.303R)(1/T1 - 1/T2). This comes from the same straight line - you are just using two points on it. NEET 2024 directly asked that Ea can be found if you know the rate constants at two different temperatures.
The slope of the Arrhenius plot (ln k vs 1/T) of a first-order reaction is -5 x 10^3 K. The value of Ea of the reaction is (Given R = 8.314 J K^-1 mol^-1):
Which of the following plots of ln k versus 1/T is consistent with the Arrhenius equation?
Activation energy of any chemical reaction can be calculated if one knows the value of:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The slope is -Ea/R. Since Ea (activation energy) and R (gas constant) are both positive, the slope is always negative. That is why the line falls from left to right.
Use Ea = -(slope) x R, with R = 8.314 J K^-1 mol^-1. The minus sign turns the negative slope into a positive Ea. Divide by 1000 at the end to convert J/mol to kJ/mol.
The intercept on the y-axis is ln A, where A is the frequency factor (pre-exponential factor). It is the value ln k would have if 1/T were extended to zero (very high temperature).
Yes. For log k (base 10) vs 1/T, the slope is -Ea/(2.303R), so Ea = -(slope) x 2.303 x R. Always read the axis label: ln k uses -Ea/R, log k uses -Ea/2.303R.
Yes. A steeper (more negative) slope means a larger Ea, so the reaction rate is more sensitive to temperature. A flatter line means a smaller Ea. If Ea = 0, the line is horizontal and k does not change with temperature.
