For a second-order reaction, [A] rightarrow products, the rate of the reaction i

Question

For a second-order reaction, [A] rightarrow products, the rate of the reaction is given as rate = [A]^2, where k is the rate constant and [A] is the concentration of reactant a. The integrated rate law for second-order reactions is 1/[A] = kt + 1/[A], where [A], is the concentration of reactant A at time t.k a is the rate constant, and [A]_0 is the initial concentration of reactant A This equation is of the type y = mx + b. Therefore, the plot of 1/[A] versus time is always a straight line with a slope k and a y intercept 1/[A]_e. Consider the second-order reaction: 2HI(g) rightarrow H_2(g) + I_2(g) Use the simulation to find the initial concentration [HI]_0 and the rate constant k for the reaction. What will be the concentration of m after t = 4.15 times 10^10 s ([HI]_t) for a reaction starting under the condition in the simulation? Express your answer in moles per liters to three significant figures.

Explanation / Answer

B)

if this is second order, then

1/A = 1/A0 + kt is valid

simply solve as follows:

1/A = 1/A0 + kt

A= 1 / (1/A0 + kt)

k --> get it from your simulation

A0 = initial concentration

t = 4.15*10^10

so

the [HI] = 1 / (1/A0 + kt)

[HI] = 1 / (1/A0 + k*(4.15*10^10)

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.