At a certain fixed temperature, the reaction A(g) + 2 B(g) ā†’ AB2(g) is found to be first order in the concentration of A and zero order in the concentration of B. The reaction rate constant is 0.0672 sāˆ’1 . If 2.000 moles of A and 4.000 moles of B are placed in a 1.000 liter container, how many seconds will elapse before the concentration of A has fallen to 0.300 mol/liter? 1. 36.6948 2. 60.226 3. 63.2373 4. 34.244 5. 49.9242 6. 23.8631 7. 51.2735 8. 29.6425 9. 36.1356 10. 28.231 Answer in units of s.

Respuesta :

Answer: 28.231 seconds

Explanation:

[tex]A(g)+2B(g)\rightarrow AB_2(g)[/tex]

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]Rate=k[A]^1[B]^0[/tex]

Thus overall order = 1

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant =[tex]0.0672s^{-1}[/tex]

t = time taken for decomposition

a = initial amount of the reactant A =[tex]\frac{moles}{volume}=\frac{2.000}{1.000L}=2.000M[/tex]

a - x = amount left after decay process = 0.300 M

[tex]t=\frac{2.303}{0.0672}\log\frac{2.000}{0.300}[/tex]

[tex]t=28.231s[/tex]

28.231 seconds will elapse before the concentration of A has fallen to 0.300 mol/liter

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