Question

For the reaction of sodium with oxygen to produce sodium oxide, the rate of disappearance of sodium is 1.2 x 10^-3 M/sec. What is the rate of appearance of sodium oxide? The rate of disappearance of oxygen? 4 Na + O₂   -> 2 Na₂O

Answer 1

4 Na + O₂   -> 2 Na₂O

-1/4 $\Delta$ [Na]/ $\Delta$ T    = - $\Delta$ [O2]/ $\Delta$ T      = 1/2 $\Delta$ [Na2O]/ $\Delta$ T

-1/4 *1.2*10^-3 = - $\Delta$ [O2]/ $\Delta$ T

3*10^-4         = - $\Delta$ [O2]/ $\Delta$ T'

The rate of disappearance of oxygen = 3*10^-4 M/sec

-1/4 $\Delta$ [Na]/ $\Delta$ T   = 1/2 $\Delta$ [Na2O]/ $\Delta$ T

-1/4*1.2*10^-3    = 1/2 $\Delta$ [Na2O]/ $\Delta$ T

3*10^-4            = 1/2 $\Delta$ [Na2O]/ $\Delta$ T

6*10^-4           = $\Delta$ [Na2O]/ $\Delta$ T

the rate of appearance of sodium oxide is 6*10^-4M/sec