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

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"