The reaction 2 NO(g) + O₂(g) → 2 NO₂ (g) is second order in NO and first order in O₂. When [NO] = 0.040 M, and 3O24 = 0.035 M, the observed rate of disappearance of NO is 9.3⨉10-5 M/s. (c) What are the units of the rate constant?

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welcome back everyone in this example, we have oxygen and carbon dioxide for the below reaction are given as first order and second order reactant, respectively. We need to determine the unit of our rate constant. If our observed rate of disappearance of our reactant, carbon dioxide is 8.9 times 10 to the fifth power polarities per second. When our carbon dioxide concentration is 100.18 moller and our concentration of oxygen is 0.17 moller. So our first step is to write out our rate law and we should recall that our rate law is expressed as our rate set equal to our rate constant K. Which to answer this prompt, we need to find units for and this is multiplied by the concentration of each of our reactant raised to their respective orders. So for our first reactant according to our equation we have carbon dioxide. So that would be times the concentration of our first reactant. Carbon dioxide. And according to the prompt, carbon dioxide is second order. So we raise it to an exploitative to here. So this is then multiplied by the concentration of our second reactant, which according to our given reaction, is oxygen. And according to the prompt, our oxygen is first order in our reaction. So we would say our concentration of 02 raised to a power of one. So we can simplify this because we don't really need to include that expletive one there because it's understood so we can say that our rate is equal to the rate constant K. Times the concentration of carbon dioxide which is a second order reactant times the concentration of oxygen, which is a first order reactant. So this is our simplified rate law. And our next step is to recall that we need our overall order for our reaction. We can find that by taking the sum of our exponents in our rate law. And so we can say that our overall order is going to be there for equal to. I'm sorry about that, It's going to be equal to two plus one, which comes from our orders of our reactant. And those are the exponents expressed in our rate law here. And so two plus one would give us three as our overall order. And so now that we have a value for the overall order, we should recall the formula to solve for our units for a rate constant and recall that that is going to be our rate constant is set equal to units of polarity raised to negative one times the value for our overall order being N. Which is then added to one. This is all in the exponent here, and then this is going to be multiplied by time raised to a power of negative one. So again we said where N is equal to the overall reaction order. And so we can say that therefore N is equal to three because we determined that our overall order four reaction is three. When we added our exponents in our rate law together. Now looking back at the prompt were given the rate of disappearance for carbon dioxide A. K. A. The rate of consumption of this reaction here As a value of 8.9 times 10 to the 5th power polarities per second. And because we know our rate constant includes time as a unit, we know that our time is going to be based on this rate of disappearance for are reacting here which is in seconds. And so we can say that based on our rate of disappearance for our reactant, carbon dioxide time is in units of seconds. So you should recognize that when we have something raised to a negative one power. That's another way of saying that we can write our rate constant so that it's equal to our molar IT e to the exponents. We have negative N plus one divided by time. So we can express it in both of these ways. And plugging in what we know for our calculation for the units of rate constant. We can say that our rate constant is equal to our polarity times negative one. Where we have our overall reaction order of three which we found above plus one multiplied by we can now say seconds because we know time is in seconds to the negative first power. And so we can simplify this so that our rate constant K. Is equal to polarity. We're going to take negative three plus one in our calculators. And recall that That's going to give us negative two in our exponents now, which is then multiplied by inverse seconds. And as we stated before, when we have something raised to a power of negative one in the exponents, we can also rewrite it as the rate constant is equal to units of Polarity to the negative 2nd power divided by seconds. And so either of these expressions here would be correct answers as the units for our rate constant. To complete this example as our final answers. So I hope that everything that I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

The reaction 2 NO(g) + O₂(g) → 2 NO₂ (g) is second order in NO and first order in O₂. When [NO] = 0.040 M, and 3O24 = 0.035 M, the observed rate of disappearance of NO is 9.3⨉10-5 M/s. (c) What are the units of the rate constant?

Verified Solution

Key Concepts

Rate Law

Order of Reaction

Units of the Rate Constant (k)

The reaction 2 NO(g) + O2(g) → 2 NO2 (g) is second order in NO and first order in O2. When [NO] = 0.040 M, and 3O24 = 0.035 M, the observed rate of disappearance of NO is 9.3⨉10-5 M/s. (c) What are the units of the rate constant?

Verified Solution

Key Concepts

Rate Law

Order of Reaction

Units of the Rate Constant (k)

The reaction 2 NO(g) + O₂(g) → 2 NO₂ (g) is second order in NO and first order in O₂. When [NO] = 0.040 M, and 3O24 = 0.035 M, the observed rate of disappearance of NO is 9.3⨉10-5 M/s. (c) What are the units of the rate constant?