<span>0.70 mol/0.250 L = 2.8 M</span>
The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
