Question

For 2H2O2---->2H2O + O2, t0.5=0.301 hr. When [H2O2] at t=0 is 0.5 M, initial rate is .......
Ans: 1.151 M/h

Answer 1

Answer:

1.1515 mol/hour

Explanation:

2H2O2--------------->2H2O + O2  (t(0.5)=0.301 hr)

From integrated rate equation for first order reaction,

$t=\frac{1}{k}ln\frac{a_{0} }{a}$

For t(0.5),

t(0.5)= 1/k x ln(a/(a/2))

0.301=1/k x ln2

Therefore, k= ln2/0.301  /hr

We know that rate(dr/dt) of a first order reaction is given by dr/dt = k[a] , where a is concentration of the reactant.

Therefore for the reaction the initial rate is

dr/dt = K [0.5]

dr/dt = ln2/0.301 [0.5]

ln2/0.301≈2.303

Therefore the initial rate of the reaction is 1.1515 mol/hour