Test #3 Q5

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Test #3 Q5

Postby nelms6678 » Thu Mar 15, 2018 12:24 pm

CHCl3(g) + Cl2(g) ---> CCl4(g) + HClg(g)
The rate of the rxn was first order with respect to chlorine and trichloromethane.

What is the rate law?

Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

Re: Test #3 Q5

Postby nelms6678 » Thu Mar 15, 2018 12:30 pm

Also Q5 asked the question:
Given the instantaneous rate of rxn is 2.54x10^-2 mol/(Lxs), and intial mass of each reactant is 1.2g confined to a 750mL vessel, what is the rate constant of this reaction?

-What equation are we even using here?

Cristina Sarmiento 1E
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Joined: Wed Nov 16, 2016 3:02 am

Re: Test #3 Q5

Postby Cristina Sarmiento 1E » Thu Mar 15, 2018 1:13 pm

Since both reactants are first order, the rate law would be
rate law = k[CHCL3][CL2].

Because you are given the instantaneous rate of reaction, 2.54 x 10^/2 mol/L and the concentrations can be solved for, you just plug those numbers into the rate law equation to find k.
To find the concentration of the reactants, divide 1.2 g by its molar mass and then divide it by 750 ml (convert it to .75L) to get the molarity of the reactants.
For CHCl3, you get 1.3 x 10^/2 mol/L and for CL2, you get 2.3 x 10^-2 mol/L.
Plugging this into the equation and solving for k, you get k = 85 s^-1

Kyle Alves 3K
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Re: Test #3 Q5

Postby Kyle Alves 3K » Thu Mar 15, 2018 10:34 pm

Did anyone get the correct answer for the other form? It did it the same way as this problem in corrections, but wanted to make sure they're both the same. Thanks!

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