The following experimental data were obtained in a study of the reaction 2 HI1g2S H21g2 + I21g2. Predict the concentration of HI that would give a rate of 1.0 * 10-5 M>s at 650 K.

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Hello. Everyone in this video being asked what concentration of H. c.. at 550 Kelvin's would give a rate of this value over here. Alright so we have to go ahead and recognize that the rate equation is equal to the K. Constant, multiplied by the concentration of H. C. L. Raised to the power of X. So we're just using this equation right over here. That's given to us in the problem. So from experiment one and 2, you see that the rate. So let's go actually write this in writing from experiments one and two. We have rate too over raped one which is equal to this K. Constant. Multiplied by the concentration of hcl raise to power of X. Over the cake constant. Multiplied by the concentration of H. C. L. Raise your power of thanks. So we're gonna go ahead and apply logarithms to both sides and use the power rule. So that gives us the log of rate to over rate one. We'll put parentheses here. This is equal to R. X. Log of the concentration of hcl two and hcl one. Alright so now basically we're solving for X over here on this right side and divide both sides by this log here. So what that does for the right side at least is that it isolates our X. Because it cancels this whole thing out and for the left we'll go ahead and simplify this. So we get that X will be equal to the log. And also of course we'll go ahead and plug in the values that we have given to us for the initial rate reactions. Alright, so log of 4.5 times 10 to the negative 5/2 100.0 times 10 to the negative five. It's all over The log of 0.60 Over 0.40. And once you put that into the calculator we see that our x value is equal to two. So now we can go ahead and write our rate creation. So rate then is equal to concentration or rather our K constant multiplied by the concentration of H. C. L. Raised to a power of two. So what we saw for earlier was this exponent here And this was an ex we saw X to be equal to two. Alright now scrolling down some more. So we know that from experiment one at 750 Calvin's the K one value is equal to rate over the concentration of hcl raise to power of two. We're just using this whole or this whole equation here. All right. And this is all equal to let's see what if you plug in the value. So rate is 2.0 times 10 to the negative five. And for the concentration of hcl there for experiment one That 0.40 and of course is raised to a power of two according to our equation. So this K one value is equal to 1.25 times 10 to the negative four units is moller to negative one times 2nd to -1. All right now from experiment three At this time 850 Kelvin's K three is equal to of course just rate over the concentration of H. C. L. Raised to a power of two. This time the right value has given to us is 6.0 times 10 to the negative four. And for the concentration that is 0.80 raised to part of two. Once you put that into the calculator received that K three is equal to 9.375 Times 10 to the -4 uses the same moller to the negative one times seconds to the -1. So we're gonna go ahead and now calculate the activating energy of the reaction which is just E. Of A. So to calculate for that that is the natural log Of K. four over K. one. And this equal to negative E. A. By are multiplied by one over T 4 -1 over T. one. So plugging in all these values then So Ellen of K four is equal to negative 106 weapons in parentheses 1060.79 kila jewels per multi that. And then our our value is 8.314 times 10 to negative three. Killah jewels per mole times kelvin's the one over T four. That's 5 50 K. And one over T. One. That is 750 K. And then we're gonna go ahead and add maybe we put this in the second line below basically going to add all this with the natural log of 1.25 times 10 to the negative four solar to power of negative one times esten power of negative one. You can see here that we did isolate this lnk of four. Alright now, continuing our simplification we can get the natural log of K. Of four is equal to, so the right side is all equal to negative 15.215107888. Go ahead and scroll down for more space. And how we isolate officially this K4 is by taking the reciprocal. So e. two the L. N. Of K. Four. Going to the E. To the power of negative 15.215107888. So what this does on the left side of this equation sign we have the E. Counseling with the natural log. So basically isolate this L. N. Or this K. Four rather so K. Four. And then on the right side once you put that into a calculator, we see that that equals to 2.47 times 10 to the neck of seven which is of course being moller two negative one times seconds two and negative one. Now lastly we're gonna go ahead and solve for the concentration of H. C. L. For the concentration of H. C. L. That is equal to Rate over K. of four, raise the power of 1/2 or just means square root. So I use this equation here to plug in the value side to have rate is 1.0 times 10 to the negative six molar per second. And from my cave for value that we just saw form that is 2.47 times 10 to the negative seven Moeller, two negative one times seconds. Two negative. One of course is all square rooted or raised a part of 1/2. So once you put that into the calculator, we can see that the concentration of H. C. L is equal to 2.0 molars. So this right here is going to be my final answer for this problem.

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Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H₃O⁺ by OH^-, which has a second-order rate constant of 1.3⨉10¹¹ M^-1 s^-1 at 25 °C. (a) If equal volumes of 2.0 M HCl and 2.0 M NaOH are mixed instantaneously, how much time is required for 99.999% of the acid to be neutralized?

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Some reactions are so rapid that they are said to be diffusion-controlled; that is, the reactants react as quickly as they can collide. An example is the neutralization of H₃O⁺ by OH^-, which has a second-order rate constant of 1.3⨉10¹¹ M^-1 s^-1 at 25 °C. (b) Under normal laboratory conditions, would you expect the rate of the acid–base neutralization to be limited by the rate of the reaction or by the speed of mixing?

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The reaction 2 NO1g2 + O21g2S 2 NO21g2 has the thirdorder rate law rate = k3NO423O24, where k = 25 M-2 s-1. Under the condition that 3NO4 = 2 3O24, the integrated rate law is 13O242 = 8 kt +113O24022 What are the concentrations of NO, O2, and NO2 after 100.0 s if the initial concentrations are 3NO4 = 0.0200 M and 3O24 = 0.0100 M?

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Values of E_a = 6.3 kJ/mol and A = 6.0⨉10⁸/(M s) have been measured for the bimolecular reaction: NO(g) + F₂(g) → NOF(g) + F(g) (d) Why does the reaction have such a low activation energy?

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A 1.50 L sample of gaseous HI having a density of 0.0101 g>cm3 is heated at 410 °C. As time passes, the HI decomposes to gaseous H2 and I2. The rate law is -Δ3HI4>Δt = k3HI42, where k = 0.031>1M ~ min2 at 410 °C. (b) What is the partial pressure of H2 after a reaction time of 8.00 h?

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Video transcript

Hello. Everyone in this video being asked what concentration of H. c.. at 550 Kelvin's would give a rate of this value over here. Alright so we have to go ahead and recognize that the rate equation is equal to the K. Constant, multiplied by the concentration of H. C. L. Raised to the power of X. So we're just using this equation right over here. That's given to us in the problem. So from experiment one and 2, you see that the rate. So let's go actually write this in writing from experiments one and two. We have rate too over raped one which is equal to this K. Constant. Multiplied by the concentration of hcl raise to power of X. Over the cake constant. Multiplied by the concentration of H. C. L. Raise your power of thanks. So we're gonna go ahead and apply logarithms to both sides and use the power rule. So that gives us the log of rate to over rate one. We'll put parentheses here. This is equal to R. X. Log of the concentration of hcl two and hcl one. Alright so now basically we're solving for X over here on this right side and divide both sides by this log here. So what that does for the right side at least is that it isolates our X. Because it cancels this whole thing out and for the left we'll go ahead and simplify this. So we get that X will be equal to the log. And also of course we'll go ahead and plug in the values that we have given to us for the initial rate reactions. Alright, so log of 4.5 times 10 to the negative 5/2 100.0 times 10 to the negative five. It's all over The log of 0.60 Over 0.40. And once you put that into the calculator we see that our x value is equal to two. So now we can go ahead and write our rate creation. So rate then is equal to concentration or rather our K constant multiplied by the concentration of H. C. L. Raised to a power of two. So what we saw for earlier was this exponent here And this was an ex we saw X to be equal to two. Alright now scrolling down some more. So we know that from experiment one at 750 Calvin's the K one value is equal to rate over the concentration of hcl raise to power of two. We're just using this whole or this whole equation here. All right. And this is all equal to let's see what if you plug in the value. So rate is 2.0 times 10 to the negative five. And for the concentration of hcl there for experiment one That 0.40 and of course is raised to a power of two according to our equation. So this K one value is equal to 1.25 times 10 to the negative four units is moller to negative one times 2nd to -1. All right now from experiment three At this time 850 Kelvin's K three is equal to of course just rate over the concentration of H. C. L. Raised to a power of two. This time the right value has given to us is 6.0 times 10 to the negative four. And for the concentration that is 0.80 raised to part of two. Once you put that into the calculator received that K three is equal to 9.375 Times 10 to the -4 uses the same moller to the negative one times seconds to the -1. So we're gonna go ahead and now calculate the activating energy of the reaction which is just E. Of A. So to calculate for that that is the natural log Of K. four over K. one. And this equal to negative E. A. By are multiplied by one over T 4 -1 over T. one. So plugging in all these values then So Ellen of K four is equal to negative 106 weapons in parentheses 1060.79 kila jewels per multi that. And then our our value is 8.314 times 10 to negative three. Killah jewels per mole times kelvin's the one over T four. That's 5 50 K. And one over T. One. That is 750 K. And then we're gonna go ahead and add maybe we put this in the second line below basically going to add all this with the natural log of 1.25 times 10 to the negative four solar to power of negative one times esten power of negative one. You can see here that we did isolate this lnk of four. Alright now, continuing our simplification we can get the natural log of K. Of four is equal to, so the right side is all equal to negative 15.215107888. Go ahead and scroll down for more space. And how we isolate officially this K4 is by taking the reciprocal. So e. two the L. N. Of K. Four. Going to the E. To the power of negative 15.215107888. So what this does on the left side of this equation sign we have the E. Counseling with the natural log. So basically isolate this L. N. Or this K. Four rather so K. Four. And then on the right side once you put that into a calculator, we see that that equals to 2.47 times 10 to the neck of seven which is of course being moller two negative one times seconds two and negative one. Now lastly we're gonna go ahead and solve for the concentration of H. C. L. For the concentration of H. C. L. That is equal to Rate over K. of four, raise the power of 1/2 or just means square root. So I use this equation here to plug in the value side to have rate is 1.0 times 10 to the negative six molar per second. And from my cave for value that we just saw form that is 2.47 times 10 to the negative seven Moeller, two negative one times seconds. Two negative. One of course is all square rooted or raised a part of 1/2. So once you put that into the calculator, we can see that the concentration of H. C. L is equal to 2.0 molars. So this right here is going to be my final answer for this problem.