Question

13.What is the half-life in minutes of a compound if 75.0 percent of a given sample decomposes in 30.0 minutes? Assume first-order kinetics.

The thermal decomposition of phosphine (PH3) into phosphorus and molecular hydrogen is a first-order reaction: 4PH3(g) → P(g) + 6H2(g) The half-life of the reaction is 35.0 s at 680°C. Calculate the first-order rate constant for the reaction: s−1 Calculate the time required for 88.0 percent of the phosphine to decompose: s

Answer #1

13. K = 2.303/t log(a/a-x)

K = 2.303/30 log(100/100-75)

=
0.0767*0.6020 = 0.0462sec^{-1}

K = 0.693/t1/2

t1/2 = 0.693/0.0462 = 15 sec

14. K = 0.693/t1/2

K = 0.693/35 = 0.0198sec^-1

. K = 2.303/t log(a/a-x)

t = 2.303/0.0198 log(100/100-88)

= 116.3*0.9208 = 107.09sec

1.Chemical kinetics
4NH3 + 3O2 = 2N2 + 6H2O
Given, rate of N2 = 0.45 Moles /Ls Calculate the rate of
NH3?
2. The thermal composition of phosphine (PH3) into phosphorus
and molecular hydrogen is a first order reaction:
4PH3(g) P4(g) + 6H2(g)
The half life of a reaction is 35 sec at 6800C. Calculate the
time required for 80% of the phosphine to decompose.

1.) The decomposition of phosphine (PH3) is a 1st
order reaction.
A.) If the half life of the reaction is 35.0s, what is the rate
constant?
B.) How long will it take for 75% of the phosphine to
decompose?

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?= min

Consider the following reaction: 2NO(g)+2H2(g)→N2(g)+2H2O(g)
What is the reaction rate at 1000 K when the concentration of NO is
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