Question:

If a reaction follows the Arrhenius equation, the plot lnK vs 1/(RT) gives a straight line with a gradient (−y) unit. The energy required to activate the reactant is:

-y unit

y unit

yR unit

y/R unit

We have, K = Ae^(-Ea/RT)
∴ lnK = ln(Ae^(-Ea/RT))
∴ lnK = lnA - Ea/(RT)
Compare it with y = mx + c
we get slope, m = -Ea/R = -y (Given)
∴ Ea = yR