1. The second order decomposition of HI has a rate constant of 1.80 x 10-3 M-1s-
ID: 1017365 • Letter: 1
Question
1. The second order decomposition of HI has a rate constant of 1.80 x 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
2. Write a nuclear equation to describe the neutron induced fission of U-235 to form Te-137 and Zr-97. Determine how many neutrons are produced in the reaction.
3. Consider the gas-phase decomposition of N2O5: N2O5(g) 2 NO2(g) + 1 half O2(g) This is known to be a first order reaction with a half-life of 102 s at 70°C. How long would it take for 35% of a sample of N2O5 to decompose at 70°C?
Explanation / Answer
1. The second order decomposition of HI has a rate constant of 1.80 x 10-3 M-1s-1. How much HI remains after 27.3 s if the initial concentration of HI is 4.78 M?
Answer =3.88 M
rate constant k =1.80 x 10^-3
t= 27.3 s
initial concentration of HI =4.78 M
plugging above values in the equation
1/[H] t =kt +1/[A]0
1/[A] t =(1.80 x 10^-3 (27.3)+1/4.78
1/[A] t =0.258
[A] t =1/0.258
= 3.8759=3.88M answer
2. Write a nuclear equation to describe the neutron induced fission of U-235 to form Te-137 and Zr-97. Determine how many neutrons are produced in the reaction.
1n0 + 235U92 ->137 Te52 +97 Zr40 + 2 1n0 +gamma rays
Answer= 2
Element
mass
U
235
Te
137
zr
97
reactants
total
mass
1
235
236
products
total
mass
137
94
234
Element
mass
U
235
Te
137
zr
97
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.