partDens = 1.5;   
diff1 = 
      0.1*(1500/300)^1.5;   
diff2 = 
      0.1*(1000/300)^1.5;   
partRadius = 
      0.05;   
atomicMass = 
      12;   
partDensMol = 
      partDens/atomicMass;
pressure = 1;   
gasConst = 
      82.06;   
temp1 = 
      1500;   
temp2 = 
      1000;   
o2Conc1 = 
      0.21*pressure/gasConst/temp1;   
o2Conc2 = 
      0.21*pressure/gasConst/temp2;   
tauTransp1 = 
      partDens*partRadius^2/(2*atomicMass*diff1*
              o2Conc1)  
kRxn1 = 1000;    
tauChem1 = 
      partDensMol*partRadius/(kRxn1*o2Conc1)
kRxn2 = 0.5;    
tauTransp2 = 
      partDens*partRadius^2/(2*atomicMass*diff2*
              o2Conc2)  
tauChem2 = partDensMol*partRadius/(kRxn2*o2Conc2)

Therefore, at 1000 K (temp2) the reaction on the surface is the RDS, whereas at 1500 K (temp1) the diffusion of O2 to the particle surface is the RDS. Note also that for FBC conditions (ca. 1000 K and 1 mm particles), the minimum residence time is of the order of an hour or so.

gasConst2 = 1.987*10^-3;
Solve[Log[kRxn1/kRxn2] == eA/gasConst2*(1/temp2 - 1/temp1), 
      eA]
eAct = %[[1, 1, 2]]

OK. 45 kcal/mol makes sense... Right?

temp3 = 2100;
temp4 = 1200;

Solve[Log[k3/kRxn1] == eAct/gasConst2*(1/temp1 - 1/temp3), 
      k3]
kRxn3 = %[[1, 1, 2]]

Solve[Log[k4/kRxn2] == eAct/gasConst2*(1/temp2 - 1/temp4), 
      k4]
kRxn4 = %[[1, 1, 2]]

diff3 = 
      0.1*(temp3/300)^1.5;   
o2Conc3 = 
      0.21*pressure/gasConst/temp3;   
tauTransp3 = 
      partDens*partRadius^2/(2*atomicMass*diff3*
              o2Conc3)  
tauChem3 = 
      partDensMol*partRadius/(kRxn3*o2Conc3)
\(diff4 = 0.1*\((temp4/300)\)^1.5;  (*cm^2/s*) 

    o2Conc4 = 
      0.21*\(pressure/gasConst\)/temp4;  (*mol/
        cm^3*) 
tauTransp4 = 
      partDens*partRadius^2/\((2*atomicMass*diff4*
              o2Conc4)\) (*seconds*) \)
tauChem4 = partDensMol*partRadius/(kRxn4*o2Conc4)

OK, at 2100 K the diffusion of O2 to the particle surface is the RDS (because the diffusion time, 69 s, is much longer than the reaction time, 67 ms. At 1200 K, reaction and diffusion times are of the same order of magnitude and complete consumption of coal occurs in several minutes. Let's now see what happens for much smaller particles, characteristic of PCC conditions.

partRadiusNew = 0.005;
tauTransp3 = partDens*partRadiusNew^2/(2*atomicMass*diff3*
              o2Conc3)  
tauChem3 = partDensMol*partRadiusNew/(kRxn3*o2Conc3)

Note also that the characteristic diffusion time is more sensitive to particle size variations than the reaction time (quadratic vs. linear relationship), whereas the reaction time is much more sensitive to temperature changes (e^x vs. x^1.5 relationship). Also, at 2100 K and for 100 mm particles, which corresponds to PCC conditions, the 100% conversion time is ca. 1 s, which makes sense. Right?

Plot[{Exp[x], x^1.5}, {x, 0, 5}];