is 0.00345 mol/L.
![k[HI]^2 ](/tpl/images/0531/9344/c44b5.png)
![[A_o]](/tpl/images/0531/9344/dc622.png)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](/tpl/images/0531/9344/8e7c2.png)
![[A_o]=5 mol/L](/tpl/images/0531/9344/ca62d.png)
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](/tpl/images/0531/9344/a4900.png)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](/tpl/images/0531/9344/3dcb7.png)
![[A]=0.00345 mol/L](/tpl/images/0531/9344/f652e.png)
Consider the second-order reaction:
2HI(g)→H2(g)+I2(g)
Use the simulation to...
Chemistry, 03.03.2020 05:56 ayoismeisalex
Consider the second-order reaction:
2HI(g)→H2(g)+I2(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 HI after t = 4.53×1010 s ([HI]t) for a reaction starting under the condition in the simulation?
Given from simulation:
Rate Law: k[HI]^2
k= 6.4 x 10^-9 l/(mol x s) at 500K
Initial Rate= 1.6 x 10^-7 mol/(l x s)
Answers: 3
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