Question

Can I please get help with these 2 questions, thank you!

G- Search in Docuiment Home Insert Design Layout References Mailings Review View & Share A chemical reaction is observed to convert Molecule A into Molecule B. The accompanying graph plots the change in concentration for both molecules against the change in time. Use this graph to determine the stoichiometry of this equation, if it is written as: Kinetics Graph for the Decomposition from A into B 1.40 1.20 1.00 Concentrat on of A 0.80 Concentrat c1 of B 0.60 0.40 0.20 0.00 10 Time (s) Select one: a·x=1/2 C. x=2 d, x=3 G- Search in Docuiment Home Insert Design Layout References Mailings Review View & Share Аавы A chemical reaction is observed to convert Molecule A into Molecule B. The accompanying graph plots the change in concentration for both molecules against the change in time. Use this graph to detarmine the stoichiometry of this equation, if it is written as: Kinetics Graph for the Decomposition from A into B 1.20 1.00 Concentratian at Concentratian at e 0.00 10 Time (s) Select one: a.x=1/2 d, x=3 Page 2 er 4 邛 + 140% 2ais acres English (Austrlia:

Answer 1

1)

The chemical reaction is of the form

A $\rightarrow$ x B

One molecule of A is converted into x molecules B

the number of molecules of B produced = ( x ) $\times$ number of molecules of A consumed

the number of molecules of B produced per unit time = ( x ) $\times$ number of molecules of A consumed per unit time

rate of production of B = - ( x $\times$ rate of consumption of A )

-rateo fproductionof B rateo fconsumptiono f A

we are going to assume the plot is linear ( it deviates slightly from the linear behaviour ) and calculate the slope of the graph determining the rate of reaction and the ratio between the rate of production of B and rate of consumption of A gives the the value of x which we will round up to nearest value

from graph

the concentration of A at time ( t₁ = 0s ) [A₀] = 1.00 M

the concentration of A at time ( t₂ = 4s ) [A₄] = 0.80 M

rate of consumption of A =

2

=

0.80- 1.00 4-0

= $\frac{-0.20}{4}$

= - 0.05 M / t

from graph

the concentration of B at time ( t₁ = 0s ) [B₀] = 0.00 M

the concentration of B at time ( t₂ = 5s ) [B₅] = 0.80 M

rate of production of B =

2

= $\frac{0.80-0.00}{5-0}$

=

0.80

= 0.16 M / t

-rateo fproductionof B rateo fconsumptiono f A

$x=\frac{-0.16 }{- 0.05}$

$x=3.2$

rounding upto nearest value x = 3

2)

The chemical reaction is of the form

A $\rightarrow$ x B

One molecule of A is converted into x molecules B

the number of molecules of B produced per unit time = ( x ) $\times$ number of molecules of A consumed per unit time

rate of production of B = - ( x $\times$ rate of consumption of A )

-rateo fproductionof B rateo fconsumptiono f A

we are going to assume the plot is linear ( it deviates slightly from the linear behaviour ) and calculate the slope of the graph determining the rate of reaction and the ratio between the rate of production of B and rate of consumption of A gives the the value of x which we will round up to nearest value

from graph

the concentration of A at time ( t₁ = 0s ) [A₀] = 1.00 M

the concentration of A at time ( t₂ = 8s ) [A₄] = 0.60 M

rate of consumption of A =

2

= $\frac{0.60-1.00}{8-0}$

= $\frac{-0.40}{8}$

= - 0.05 M / t

from graph

the concentration of B at time ( t₁ = 0s ) [B₀] = 0.00 M

the concentration of B at time ( t₂ = 6s ) [B₅] = 0.60 M

rate of production of B =

2

=

0.60-0.00 6-0

= $\frac{0.60}{6}$

= 0.10 M / t

-rateo fproductionof B rateo fconsumptiono f A

$x=\frac{- 0.10 }{-0.05}$

$x=2.0$

therefore x = 2