When 442 mg of NO2 is confined to a 150. mL reaction vessel and heated to 300°C, it decomposes by a second-order process. In the rate law for the decomposition of NO2, k = 0.54 1/(M·s).

Question

When 442 mg of NO2 is confined to a 150. mL reaction vessel and heated to 300°C, it decomposes by a second-order process. In the rate law for the decomposition of NO2, k = 0.54 1/(M·s).

A) what is the initial reaction rate?
B) what is the reaction rate if the mass of NO2 increased to 728mg?

I typically always put down my work...but I have no idea where to start with this other than converting NO2 to moles? completely lost here...

Answer 1

To answer these questions, we need to apply the principles of chemical kinetics and the concepts of concentration, rate laws, and reaction order.

Let's break down the problem step by step:

Step 1: Convert mass of NO2 to moles
To begin, you correctly suggested converting the mass of NO2 to moles. To do this, we can use the molar mass of NO2.

Molar mass of NO2 = 46 g/mol (N: 14.01 g/mol, O: 16.00 g/mol)

So, we can convert the given masses of NO2 to moles:
1. For the first case:
Mass of NO2 = 442 mg = 0.442 g
Moles of NO2 = (0.442 g) / (46 g/mol) = 0.0096 mol

2. For the second case:
Mass of NO2 = 728 mg = 0.728 g
Moles of NO2 = (0.728 g) / (46 g/mol) = 0.0158 mol

Now we have the number of moles of NO2 in both cases.

Step 2: Calculate the initial reaction rate (A)
The initial reaction rate, denoted as "A," can be obtained from the rate law equation for a second-order process. The rate law equation for the decomposition of NO2 is given by:

Rate = k * [NO2]^2

Here, k represents the rate constant, and [NO2] denotes the concentration of NO2.

In this case, since the reaction vessel volume remains constant at 150 mL, we can assume the concentration of NO2 at the start is the same as in the reaction vessel.

For the first case:
[NO2] = moles of NO2 / volume of reaction vessel
[NO2] = 0.0096 mol / 0.150 L = 0.064 M

Substituting the values into the rate law equation:
Rate = k * [NO2]^2
Rate = 0.54 1/(M·s) * (0.064 M)^2

Calculating this expression will give us the initial reaction rate (A).

Step 3: Calculate the reaction rate for the increased mass of NO2 (B)
To find the reaction rate in the second case, we follow a similar procedure.

[NO2] = moles of NO2 / volume of reaction vessel
[NO2] = 0.0158 mol / 0.150 L = 0.105 M

Substituting the values into the rate law equation:
Rate = k * [NO2]^2
Rate = 0.54 1/(M·s) * (0.105 M)^2

Calculating this expression will give us the reaction rate in the second case (B).

So, by following these steps, you can find both the initial reaction rate (A) and the reaction rate when the mass of NO2 is increased (B).

Answer 2

You know it's second order so your rate=k[NO2]^2

All you do is like you said, find moles, use that to find molarity (convert mL to L) and then plug the concentration into the equation. You're given the k.

Same thing with part B, just changing the concentration :)