A rate is equal to 0.0200 m/s. if [a] = 0.100 m and rate = k[a]2[b]2, what is the new rate if the concentration of [a] is increased to 0.200 m?

The rate equation is r = k[a]²[b]².

Take logarithms of both sides.

ln r = ln k + 2 ln [a] + 2 ln [b]

When [a] = 0.0.100 m, r₁ = 0.0200 m/s.

Therefore

ln (0.0200) = ln k + 2 ln (0.0100) + 2 ln [b] (1)

When a = 0.200 m, obtain the new rate, r₂, as

ln r₂ = ln k + 2 ln (0.0200) + 2 ln [b] (2)

Subtract (1) from (2) to obtain

ln r₂ - ln (0.0200) = 2{ln (0.0200 - ln (0.0100) }

ln (r₂/0.0200) = 2 ln (0.0200/0.0100)

= 2 ln (2)

= ln (2²) = ln (4)

Therefore

r₂/0.0200 = 4

r₂ = 0.0800 m/s

Answer: 0.0800 m/s