When 0.13 g of H2 and 0.18 g of I2 are confined to a 150. mL reaction vessel and heated to 700. K, they react by a second-order process (first order in each reactant), with k = 0.063 L·mol-1s-1 in the rate law (for the rate of formation of HI).
(a) What is the initial reaction rate?
(b) By what factor does the reaction rate increase if the concentration of H2 present in the mixture is tripled?
H2 + I2 ==> 2HI
rate = k(H2)(I2)
you know k, you can calculate (H2) and (I2) from PV = nRT
For part b just triple the (H2) calculated in part a and recalculate rate
To determine the initial reaction rate in this second-order process, you need to use the given rate law and the initial concentrations of the reactants.
(a) The rate law for the formation of HI is given as:
rate = k[H2][I2]
In this case, the concentrations of H2 and I2 at the start of the reaction are given as 0.13 g and 0.18 g, respectively. To convert these masses to molar concentrations, you need to use the molar mass of each compound.
The molar mass of H2 is 2 g/mol, so the moles of H2 present initially can be calculated as:
moles of H2 = mass of H2 / molar mass of H2 = 0.13 g / 2 g/mol = 0.065 mol
Similarly, using the molar mass of I2 as 254 g/mol, you can calculate the moles of I2 present initially as:
moles of I2 = mass of I2 / molar mass of I2 = 0.18 g / 254 g/mol = 0.000708 mol
Now, you can substitute these values into the rate law:
rate = k[H2][I2] = 0.063 L·mol-1s-1 * (0.065 mol) * (0.000708 mol)
Note that the reactant concentrations need to be expressed in liters because the rate constant has units in L·mol-1s-1.
Simplifying the calculation:
rate = 0.063 L·mol-1s-1 * 0.065 mol * 0.000708 mol
= 2.52 x 10^-6 mol·L-1s-1
Therefore, the initial reaction rate is 2.52 x 10^-6 mol·L-1s-1.
(b) To determine the factor by which the reaction rate increases if the concentration of H2 is tripled, you can use the rate law and compare the initial rate (with the original H2 concentration) to the new rate (with the tripled H2 concentration).
Let's denote the initial concentration of H2 as [H2]1 and the concentration after tripling it as [H2]2.
Comparing the two rates:
rate1 = k[H2]1[I2]
rate2 = k[H2]2[I2]
We already know the initial rate, so let's calculate the new rate.
Triple the concentration of H2:
[H2]2 = 3 * [H2]1
Now, substitute these values into the rate equation:
rate2 = k[H2]2[I2] = k * 3[H2]1[I2]
Divide the new rate (rate2) with the initial rate (rate1) to find the factor by which the reaction rate increases:
factor = rate2 / rate1 = (k * 3[H2]1[I2]) / (k[H2]1[I2])
The [I2] term cancels out, and you are left with:
factor = (k * 3[H2]1) / (k[H2]1) = 3
Therefore, if the concentration of H2 is tripled, the reaction rate will increase by a factor of 3.