Question

a.) Calculate the partial pressure in bar of phosphorus pentachloride produced in a system in which phosphorus trichloride and chlorine gases are initially each at a partial pressure of 3.00 bar and allowed to reach equilibrium. The reaction is run at a temperature in which the equilibrium constant = 1.12. Note: This K value is for the reaction written as: PCl3(g) + Cl2(g) ----> PCl5(g)

b.) Strontium-90 is a radionuclide that is a legacy of nuclear weapons development and testing. It is particularly dangerous because it takes the place of calcium in bone marrow, and has a half life of 28.1 years. Suppose 1.20 micrograms were absorbed by a human. How many micrograms will remain after 34.0 years?

c.) The chemical decomposition of reactant species A is followed in a spectrophotometer to determine its decreasing concentration with time. The following data is taken:

time(min)      [A] mmol/L
---------------------------
  0              90.0            
  10             66.58            
  20             49.26            
  30             36.44            
  40             26.96            
---------------------------

Determine the order of the reaction and the rate constant. If the reaction turns out to be first order, report its half-life in minutes. If it turns out to be second order, report its rate constant in L/mmol-min.

Answer 1

a) Given equilibrium is PCl Cl2 <--->PCI K 1.12 Let x be the change in pressure of each gas at equilibrium PCI 3.00 PCI 2 5 0.0 Initial Change Equilibrium 3.00 3.00 -x 3.00-х 1.12 (3-x)(3-x) 1.12 x2-6x +9 x 1.12x6.72x +10.08 1.12x2-7.72x +10.08-0 By solving, x 5.142857 and 1.75 x cannot be 5.142857, as it is more than initial value. So,x-1.75 Partial pressure of PCl,-1.75 bar

b) Half life of Sr-90 is 28.1 years We know that all radio active substances follow first order kinetics, The activity constant is given by 2.303 N log whereX- activity constant No- initial conc. of radio active element N Remaining conc. of radio active element after time t We know that the half life and activity constant are relatedas 0.693 -0.024662 years a t,28.1 years 12 Given that initial amount absorbed by body No-1.20 ug time 34.0 years 2.303, N 0 N Axt 0.024662 year x34.0 N2.303 log years 0.364093 2.303 No 1036409s2.312558 1.20μ 0.518906 μg 0 2.312558 2.312558 Amount of Sr-90 remaining 0.52 μg

time [A] InA 1/[A] C) 10 20 30 40 90 66.584.198 0.01502 49.263.897 0.0203 36.44 3.596 0.027442 26.963.294 0.037092 Test for zero order: [A] vs time 100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 0

Test for First order: In[A] vs time 5.000 4.500 4.000 3.500 3.000 2.500 2.000 1.500 1.000 0.500 0.000 yー-0.030x + 4.499 10 20 30 40 50 Test for second order: 1/[A] vs time 0.04 0.035 0.03 0.025 0.02 0.015 0.01 0.005 0 10 20 30 40 50 0

First order rate equation is A straight line will be obtained if we plot a graph between In[A] on Y-axis and time on X-axis. A straight line is obtained for the graph between In[A] vs time, so the order of the reaction is 1 (ONE) Negative Slope of the line is the rate constant From graph, The equation for the straight line is y -0.030x+ 4.499 So the rate constant- slope (-0.030) 0.030 min-1. Half life for the reaction 0.693/k 0.693/0.030 min-1 23.1 min