. For 2A + 3B → 2C, [A]0 = 0.15, [B]0 = 0.12, and the initial rate is 0.10 mol/Ls. Assume that the reaction described above is first order in both reactants A and B. Calculate the value of the reaction rate constant.

The rate constant k is ;

K = 2572 L^4/mol^-4•s

Explanation:

In this question, we are asked to calculate the rate constant given the chemical equation above.

First, we need to write a complete expression for the rate law. The reaction is a second order type.

We write the rate law as follows;

Rate = k * [A]^2 * [B]^3

Where k is the rate constant and the square brackets indicate the concentrations of both A and B. Let’s input the values :;

0.1 = k * (0.15)^2 * (0.12)^3

0.1 = k * 0.0225 * 0.001728

0.1 = k * 0.00003888

K = 0.1/0.00003888

K = 2572 L^4/mol^-4•s

i had this on my unit activity, i can you if you are a plato student. are you a plato student?

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied