a)
rate = -(1/2)d[H2]/dt
rate = - d[O2]/dt
rate = (1/2)d[H2O]/dt
b)
we have,
rate = - d[O2]/dt = (1/2)d[H2O]/dt
- d[O2]/dt = (1/2)d[H2O]/dt
- (-0.23 mol/L.s) = (1/2)*d[H2O]/dt
d[H2O]/dt = 0.46 mol/L.s
Answer:
H2O is increasing at the rate of 0.46 mol/L.s
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