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18 votes
Hi! I am struggling with this problem and if anybody can correctly answer this problem I would gladly appreciate it, thanks!

Hi! I am struggling with this problem and if anybody can correctly answer this problem-example-1
User Dequis
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2 Answers

20 votes
20 votes

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User Voulzy
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13 votes
13 votes

Answer:

43.2 x 10⁻¹¹

Step-by-step explanation:

The rate law equation for the reaction is:

Rate = k[CH₃CHO]²

In the table, you are given some of the concentrations and rates. What you are not given is the value of "k". Since you are only given the concentration in Trial 3, you need to know the value of "k" in order to find the rate. To find this value, plug the information from one of the trials into the equation and simplify to find "k".

Rate = k[CH₃CHO]²

2.70 x 10⁻¹¹ = k[2.00 x 10⁻³]²

2.70 x 10⁻¹¹ = k[4.00 x 10⁻³]

6.75 x 10⁻⁶ = k

Now that you know the value of "k", you can use it and the concentration from Trial 3 to find the rate.

Rate = k[CH₃CHO]²

Rate = (6.75 x 10⁻⁶)[8.00 x 10⁻³]²

Rate = (6.75 x 10⁻⁶)[6.4 x 10⁻⁵]

Rate = 4.32 x 10⁻¹⁰ -------> 43.2 x 10⁻¹¹

User UndergroundFox
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