Consider the reaction data.

A⟶products

T () (−1)
225 0.394
675 0.812
What two points should be plotted to graphically determine the activation energy of this reaction? To avoid rounding errors, use at least three significant figures in all values.

x1=

y1=

x2=

y2=

Determine the rise, run, and slope of the line formed by these points.

rise:

run:

slope:

What is the activation energy of this reaction?

(a) x₁= 0.004 444; (b) y₁ = -0.9545; (c) x₂ = 0.001 905; (d) y₂ = -0.4541;

(e) rise = 0.5004; (f) run = -0.002 539; (g) slope = -197.1; (h) Eₐ = -1.64 kJ·mol⁻¹

Explanation:

This is an example of the Arrhenius equation:

Thus, if you plot ln k vs 1/T, you should get a straight line with slope = -Eₐ/R and a y-intercept = lnA

(a) x₁

x₁= 1/T₁ = 1/225 = 0.004 444

(b) y₁

y₁ = ln(k₁) = ln0.385 = -0.9545

(c) x₂

x₂= 1/T₂ = 1/525 = 0.001 905

(d) y₂

y₂ = ln(k₂) = ln0.635 = -0.4541

(e) Rise

Δy = y₂ - y₁ = -0.4541 - (-0.9545) = -0.4541 + 0.9545 = 0.5004

(f) Run

Δx = x₂ - x₁ = 0.001 905 - 0.004 444  = -0.002 539

(g) Slope

Δy/Δx = 0.5004/(-0.002 539) K⁻¹ = -197.1

(h) Activation energy

Slope = -Eₐ/R

Eₐ = -R × slope = -8.314 J·K⁻¹mol⁻¹ × (-197.1 K⁻¹) = 1638 J/mol = 1.64 kJ/mol

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it

b. examine and describe cells