The irreversible gas-phase reaction A +B → C + D is carried out in an isothermal packed bed reactor. The feed is equimolar, with CA0 = CB0 = 2 mol/L. The initial volumetric flow rate is 4 L/s. The reaction is first order with respect to A and first order with respect to B, and the rate constant is 0.01 L mol-1 s-1. The pressure drop parameter, α, is 0.0015 kg-1. Starting from the differential form of the design equation, what is the percent conversion of A, expressed to one decimal place. if 400 kg of catalyst are used in the reactor? Note that this problem may be solved either using a math solver (Matlab, Polymath, etc) or by hand. You are encouraged to try both methods.

The answer is a 0.64 volts

    m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| =  √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| =  √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| =  14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s