CHM 2046C Module Nine Sample Exam Answers

Part F: Concentration-Time Relationships Half Life 1^st Order 10 pts

Use the following t_½ = 0.693/k to solve:

Sucrose, C₁₂H₂₂O₁₁, decomposes to fructose and glucose in acid solution with the rate law:

Rate = k[C₁₂H₂₂O₁₁] k = 0.208 h^-1 at 25 ^oC

What amount of time is required for 75.0% of the initial concentration of sucrose to decompose?

t_½ = 0.693/0.208 hr = 3.33 hr

at Time_zero the amount remaining = 100%

after one half life interval (3.33 hr) the amount remaining = 50%

the amount of substance that has decomposed = 50%

after a second half life interval of time that has elapsed

(3.33x2= 6.66 hr)

The amount remaining is ½ of 50% or 25%

The amount of substance that has decomposed is 100%-25% = 75%

Answer 6.66 hr

87.5%?

after a third half life interval of time that has elapsed

(3.33x3= 9.99 hr)

The amount remaining is ½ of 25% or 12.5%

The amount of substance that has decomposed is

100%-12.5% = 87.5%

Answer 9.99 hr

93.75% ?

after a fourth half life interval of time that has elapsed

(3.33x4= 13.32 hr)

The amount remaining is ½ of 12.5% or 6.25%

The amount of substance that has decomposed is

100%-6.25% = 93.75%

Answer 13.32 hr

Read Pages 15.4 712-716. Addition Problems: #7-10