Question #706ce

1 Answer
May 1, 2017

Answer:

The answer is (E) #8#.

Explanation:

You know that

#color(blue)("rate" = k * ["NO"_2]^2 * ["Cl"_2])#

You now double the concentrations of the two reactants

#["NO"_ 2]_ "new" = color(red)(2) * ["NO"_2]#

#["Cl"_ 2]_ "new" = color(red)(2) * ["Cl"_2]#

The rate of the reaction will now be

#"rate"_ "new" = k * ["NO"_ 2]_ "new" * ["Cl"_ 2]_ "new"#

This is equivalent to

#"rate"_ "new" = k * (color(red)(2) * ["NO"_2])^2 * (color(red)(2) * ["Cl"_2])#

#"rate"_ "new" = k * color(red)(2)^2 * color(red)(2) * ["NO"_2] * ["Cl"_2]#

#"rate"_ "new" = 8 * color(blue)(k * ["NO"_2] * ["Cl"_2])#

This means that you have

#"rate"_ "new" = 8 * color(blue)("rate")#

The rate of the reaction will increase by a factor of #8#.