Molecular reaction rates are often determined in part by how much energy is made

Question

Molecular reaction rates are often determined in part by how much energy is made available in a collision. This in turn, is determined in large part by momentum conservation. For this problem we will consider how much energy isolated interacting molecules must absorb as result of constraints created by momentum conservation. One way ozone is created in the upper atmosphere is by monatomic oxygen combining with molecular oxygen: O + O_2 rightarrow O_3. (Atomic oxygen atoms are produced in the upper atmosphere by ultraviolet light breaking up O_2 molecules.) For this problem, take the x-axis to be horizontal with positive to the right. O has a mass of 16D, O_2 of 32D, and O_3 of 48D. Suppose an O atom traveling with a speed of 200 m/s collides with a stationary O_2 molecule and they bind into an ozone (O_3) molecule. What will be the velocity of the combination? Explain your reasoning. Calculate the gain or loss of kinetic energy (1/2mv^2). Express a gain as a positive number and a loss as a negative number. Use units of D-m^2/s^2. Now suppose the situation is reversed: an O_2 molecule traveling with a speed of 200 m/s collides with a stationary O atom and they bind into an ozone (O_3) molecule. What will be the velocity of the combination? Explain your reasoning. While these initial configurations are possible, they are both unlikely. It is unlikely to find a molecule at rest. If the O atoms and the O_2 molecules have the same temperature, and the average speed of an O atom is v_0, what would the average speed of an O_2 molecule be? Explain your reasoning.

Explanation / Answer

5.1

According to question momentum of the system is conserved both before and after collision

Before collision Oxygen atom is moving with velocity 200 m/s but Oxygen molecule is at rest. Hence momentum before collision is 16D*200+32D*0= 3200 D.m/s

Let's consider After collision Ozone (O3) is moving with speed V. So momentum after collision is 48D*V

Now equating [momentum before collision = momentum after collision]

3200 D.m/s=48D*V So V= 3200/48= 66.66 m/s

Here we get less velocity as the mass of the system increases after collision and the energy of Oxygen atom utilised to move Oxygen molecule at rest.

5.2

Change in Kinetic Energy (k.E.) = Final K.E. -Initial K.E.= (1/2)48D*66.662 - (1/2)16D*2002 = -213354.66 D.m2/s2

So the Kinetic Energy is lost in this system

5.3

When the situation is reverse as per the question

Momentum is conserved Initial Momentum = Final Momentum

16D*0+32D*200=48D*V So we get V= (32*200)/48 = 133.33 m/s

5.4

The average K.E. of a molecule is directly propertional to the temperature. Higher the temperature faster the molecule moves. We know the Ideal gas equation is PV=nRT average K.E. (1/2 mVx2 ) average = (3/2) kT where 'k' is Boltzman's constant which is equivalent to (R/N) where R is universal gas constant and N is Avogadro's No.

From the above equation we can calculate (v2)average = 3kT/m = 3RT/mN = 3RT/M (where M is molecular mass)

According to the problem temperature is constant for both the cases Hence only dependent quantity is molecular mass so we can say  (v2)average is inversly propertional to M (Molecular mass)

Hence (vo/voxygen molecule)2 = 32D/16D So Voxygen molecole = vo/2 = 0.707 vo

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