In a first-order decomposition reaction. 20.8% of a compound decomposes in 7.8 min. how long (in min) does it take for 88.2% of the compound to decompose?

t = 71.47 min

Explanation:

Using integrated rate law for first order kinetics as:

Where,  

is the concentration at time t

is the initial concentration

Given:

20.8 % is decomposed which means that 0.208 of is decomposed. So,

= 1 - 0.208 = 0.792

t = 7.8 min

k = 0.0299 min⁻¹

Also,

Given:

88.2 % is decomposed which means that 0.882 of is decomposed. So,

= 1 - 0.882 = 0.118

t = ?

t = 71.47 min

i just learnt this recently but i need to look through my notes again

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.