J.R. S. answered 11/27/19
Ph.D. University Professor with 10+ years Tutoring Experience
1). CH4(g) + 2O2(g) ==> CO2(g) + 2H2O(g)
rate = -d[CH4]/dt
rate = -1/2d[O2]/dt
rate = +d[CO2]/dt
rate = +1/2d[H2O]/dt
2). N2(g) + 3H2(g) ===> 2NH3(g)
a) For every 3 moles/molecules of H2 that disappears, 2 moles/molecules of NH3 appears
rate = -1/3 ∆[H2]/∆t = +1/2∆[NH3]/∆t
-1/3(0.074 M/s) = +1/2 (x)
0.0247 M/s = x/2
x = 0.0493 M/s = rate of formation of NH3
b) For every 3 moles/molecules of H2 that disappears, 1 mole/molecule of N2 disappears. So it is at a rate 1/3 that of H2.
rate = -1/3 ∆[H2]/∆t = -∆[N2]/∆t
-1/3(0.074 M/s) = -x
0.0247 M/s = x = rate of disappearance of N2