At a certain temperature the concentration of NO was 0.400 M and that of Br2 was 0.255M. At equilibrium the concentration of NOBr was found to be 0.250 M. What is the value of Kc at this temperature?
Note: I was unable to attain the correct answer for this question, if using this as a model to answer a similar question take my advice here with a grain of salt.
Since you are dealing with a homogenous solution, in this case everything is a gas, that means things are much more simple. The equation that we will be working with is for the equilibrium constant Kc.
You are given the concentration of your product at equilibrium temperature which can be used directly in the equation. But you still need to find the other two variables. The problem is that the beginning of the problem does not explicitly state that the mixture is at equilibrium. As such it could not be at equilibrium at all (just the initial conditions) or it could be at equilibrium in which case we are not given what the equilibrium concentration is. Since the second scenario makes things more difficult we will go with the initial scenario. In this case the amount of free Br2 remaining is 0.255 minus half the molar concentration of product (due to the equation for the reaction). The amount of nitrogen monoxide remaining is equal to 0.400 minus the mols of nitrosyl bromide at equilibrium.
Mol Br2 =
0.255 mol - 0.5(0.250) = 0.13 mol
So your equilibrium constant will be 21.37 at the temperature given for equilibrium.
Moore, John; Stanitski, Conrad; Jurs, Peter. Chemistry - The Molecular Science. 1st ed. USA, Harcourt. 2002
The question had two parts.. but i forgot to include the second
portion of it...
At this temperature the rate constant for the reverse reaction is 370 /M2s. What is the rate of the reaction when the initial concentration of NO is 0.400M and that of Br2 is 0.255M ?
This information will definitely allow me to get the answer you seek. First off, determine the rate law. To do this find an example above where the concentration of one of your reactants stays the same. Then look at the impact on the rate.
Experiment 2 Rate / Experiment 1 Rate = 96(
So, what changed between experiment number 2 and 1? The concentration of nitrogen monoxide doubled, but the rate quadrupled, went to the second power. This shows that for the rate law the mols of nitrogen monoxide affect the rate to the square. Now for the nitrogen monoxide remaining constant and the bromine changing.
Experiment 3 Rate / Experiment 2 Rate = 192(
The amount of bromine doubled and the rate doubled meaning that this affects the rate directly, now we can write out the rate law with the proper exponents:
Rate = k [NO]2 [Br2]
Finally with the rate in place we can determine the rate constant by just rearranging the above formula and plugging in any of the information we were given.
k = Rate / ([NO]2 [Br2])
k = 24 (M/s) / ([0.150 (M)]2 [0.240 (M)])
k = 24 (
k = 4444.44 /s M2
Now with the value of the rate constant determined we can use the same rate equation to determine the initial rate using the conditions outlined in the problem.
Rate = 4444.44 (/s
So, we now have our rate, and if you look at the rates we were given in the problem for different concentrations this rate is in line with some of the rates at higher concentrations.
Moore, John; Stanitski, Conrad; Jurs, Peter. Chemistry
- The Molecular Science. 1st ed. USA, Harcourt. 2002
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