Nitrogen Monoxide and Oxygen reaction?

Question

The reaction rate is given r = k [NO]2 x [O2] ([NO]2 = where 2 is square root)

where r = reation rate and k = reaction coefficient, which increases with increasing temperature.

(a) What is the increase int he reation rate when the concentration of Nitrogen oxide is Doubled and the Oxygen is Tripled?

(b) How much do you need to increase the nitrogen concentration to affect a nine-fold increase in reaction rate, without changing other factors that affect the reation rate?

P.S. I would like to know how to type square and cube, same for square root and cube root

Answer 1

(a) Since rate = k [NO]^(1/2) [O2], then

new rate = k * 2^(1/2) * 3, or 4.24 times as fast.

(b) Solve for [NO]:

[NO] = (rate / k * [O2])^2

Substituting, [NO] = (9/1)^2 = 81 times [NO] to get 9x reaction rate with no change in [O2].

As you probably have guessed, the accepted symbol for exponentiation is the carat (^). So "x squared" is x^2 and "x cubed" is x^3. Since radicals are really fractional exponents, then "square root of x" is x^(1/2), "cube root of x" is x^(1/3). "The cube root of x squared" would be x^(2/3).