A rate is equal to 0.0200M/s. If [A]=0.100 M and rate= k[A]^1, what is the new rate if the concentration of [A] is increased to 0.200M?

The new rate if the concentration of [A] increased to 0.200 M is 0.0800 m/s.

Explanation:

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

The new rate is 0.0800 m/s.

i think the answer is terrae but not 100 percent sure. ▓▬▬ §

the answers would be

the first one and the second one