Question

Question #8a4f6

Answer 1

The first thing you have to do is determine the concentrations of hydrogen gas and iodine gas present in the reaction vessel.

You go about doing this by using their respective molar masses to determine how many moles of each are present, then using the number of moles and the volume of the vessel to calculate their respective concentrations.

`"0.52 g" * "1 mole H"_2/"2.02 g" = "02574 moles H"_2`

`C_("hydrogen gas") = n_(H_2)/V = "0.2574 moles"/(750 * 10^(-3)"L") = "0.343 M"`

`"0.19 g" * "1 mole I"_2/"253.81 g" = "0.0007486 moles I"_2`

`C_("iodine gas") = n_(I_2)/V = "0.0007486 moles"/(750 * 10^(-3)"L") = "0.000998 M"`

Since your reaction is second-order overall, and first-order in both reactants, the expression for your rate of reaction will be

`"rate" = k * [H_2] * [I_2]`

Plug in your values to solve for the initial rate of reaction

`"rate" = "0.063 L" * "mol"^(-1) * "s"^(-1) * "0.343 mol" * "L"^(-1) * "0.000998 mol" * "L"^(-1)`

`"rate" = "0.0000261 mol" * "L"^(-1) * "s"^(-1)`, or

`"rate" = 2.61 * 10^(-5) "mol" * "L"^(-1) * "s"^(-1)"`

SIDE NOTE The units of your rate constant were wrong. For a second-order reaction, k is expressed in `"M"^(-1) * "s"^(-1)`, which is equivalent to `"mol"^(-1) * "L" * "s"^(-1)`.

Now for point b). Because your reaction is first-order with respect to both reactants, when one of the concentrations doubles, the rate will double as well.

In this case, the concentration of hydrogen gas would go from 0.343 M to 0.686 M, which would make the rate of reaction to double. So, the rate of reaction increases by a factor of 2.