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evablogger [386]

3 years ago

6

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final

concentration of A at the end of this same interval if its concentration was initially 1.600 M?

1 answer:

Serhud [2]3 years ago

5 0

Complete question:

Consider the hypothetical reaction 4A + 2B → C + 3D

Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?

Answer:

the final concentration of A is 0.992 M.

Explanation:

Given;

time of reaction, t = 4.0 s

rate of change of the concentration of B = -0.0760 M/s

initial concentration of A = 1.600 M

⇒Determine the rate of change of the concentration of A.

From the given reaction: 4A + 2B → C + 3D

2 moles of B ---------------> 4 moles of A

-0.0760 M/s of B -----------> x

$x = \frac{4(-0.076)}{2} \\\\x = -0.152 \ M/s$

⇒Determine the change in concentration of A after 4s;

ΔA = -0.152 M/s x 4s

ΔA = -0.608 M

⇒ Determine the final concentration of A after 4s

A = A₀ + ΔA

A = 1.6 M + (-0.608 M)

A = 1.6 M - 0.608 M

A = 0.992 M

Therefore, the final concentration of A is 0.992 M.

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The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

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