Consider the reaction: 8 H 2 S(g) + 4 O 2 (g) → 8 H 2 O(g) + S 8 (g) Complete the table.

well everyone's on this problem trying to refer to this reaction right over here. I'm being told to provide the missing information given below here. So we need to find these values here. But this over here is given. So let's recall that if we have a reaction and it's equal to little A capital A. That yields floor B. Capital B. The rate of the reaction here is an equal to negative one little A divided by the concentration of the change of concentration of A. Divided by delta T. So change in time and this is equal to one over little. Be multiplied by the change in concentration of big B. Over delta T. So are given reaction. Like we said it's just right over here. Let's go ahead and apply this to our formula. So the rate then is equal to negative 1/4, Notified by the change or change in concentration of NH three divided by Delta T. This is equal to negative 1/5. Multiplied by the change in concentration of 02 divided by delta T. But it's also equal to negative or just not negative but positive. 1/4 will divide by the train of concentration of R. N. O. Divided by delta T. Which is equal to 1/6. Raised to the power of delta H. 2 0. Divided by Delta Team. So we have the delta. So the change in concentration of N. H. Three divided by delta T. Is equal to negative 0.48 m per second. Alright now starting off with this right over here. So this is just a change of concentration of R. 02, divide by change in temperature, scroll down from our space and we can go back up and right in front of answers then. All right, so we have negative 1/5. Multiplied by the change of concentration of R. 02 divided by our delta T. Is equal to negative 1/4. Multiplied by the change in concentration of R. N. H. Three divided by delta T. We're gonna go ahead and simplify this and adding in the non concentration of R. NH three. So that's negative 1/5. Multiplied by delta 02 divided by delta T. You're going to negative 1/4. Multiplied By negative 0. molars per second. Simplifying this further rehab negative 1/5. To buy by delta oh two divided by delta T. Equaling two. We get 0.12 molar per second. On the right side of our equation. They finally isolating our delta oh two divided by our delta T. Is equal to 0.12 molars per second, multiplied by negative five. We finally get that delta 02 divided by delta T. Is equal to negative 0.60 molar per second. Let's go ahead and fill this out in our little chart here. So again this is negative 0.60 molars per second. And moving on to delta N. O divided by delta T. Let's maybe do this in a different color. I gave her something for delta N. O. Divided by delta T. Alright let's go ahead and do this. So for this one here we have 1/4. Multiplied by the change in concentration of R. N. O. Divided by the concentration. Already changing early change in time rather. And this is equal to negative 1/4. Delta and H. Three divide by delta T. That plugging in the known value here we have 1/4, multiplied by delta N. O. Divided by delta T. Is equal to negative 1/4. Multiplied by negative 0.48 molars per second. We're gonna go ahead and combine and simplify the right side of this equation. So on the left side will just remain the same or just basically copying over everything. Alright, so then this is equal to 0. molars per second. Now isolating the Delta and Oh divided by Delta T. recently multiply both sides by four. So multiplying everything here by four, basically what we get that is that the delta N. O. Divided by delta T. Is equal to 0.48 molars per second. So let's go ahead and scroll up and fill that out. So again this missing space here is equal to 0. molars per second. Alright, we're gonna go ahead and scroll down and move on to our second here. So now we're gonna go ahead and solve for the delta H. 20. Divided by dr team. Alright so this is delta H. 20 divided by delta T. Is equal to negative 1/4 delta And age three over the Delta T. So change in time. So we're gonna go ahead and fill out the values that we do know again. So we just have 16 multiplied by delta H. 20 over delta T. Equaling two negative 1/4, multiplied by negative 0.48 molars per second. Go ahead and combine what's on the right side here. So we have 1/6 delta H. 20 over delta T. Equaling to 0.012 molars per second. We want to go ahead and isolate what we're trying to find which is this over here. So we just multiply both sides by six. Once we do so we get that delta Or the change in concentration of H20 divided by the change in temperature is unequal to 0.072 molars per second. Again, we can go ahead and scroll back up and plug in this value. That is 0.072 molars per second. And now finally calculating for a rate. Alright, so for our rates here this is equal to negative 1/4, multiplied by delta and H three, divide by delta team. And we get the rate is equal to negative 1/4, multiplied by negative 0.048 molars per second. Which then gives us that the rate is equal to 0. molars per second. I want to go ahead and scroll up and fill that out. So they said that the rate is an equal to 0. molars per second. So all of these over here are going to be my final answers for this problem. Thank you so much for watching.

Ch.15 - Chemical Kinetics

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Tro 5th Edition

Ch.15 - Chemical Kinetics

Problem 30

Chapter 15, Problem 30

Consider the reaction: 8 H₂S(g) + 4 O₂(g) → 8 H₂O(g) + S₈(g) Complete the table.

Verified step by step guidance

Step 1: Identify the balanced chemical equation. In this case, it is 8 H2S(g) + 4 O2(g) → 8 H2O(g) + S8(g). This equation tells us that 8 moles of hydrogen sulfide react with 4 moles of oxygen to produce 8 moles of water and 1 mole of sulfur.

Step 2: Identify the given information in the problem. This could be the initial amounts of reactants, the amount of one of the products, or the final amounts of the reactants. This information will be used to fill in the table.

Step 3: Use the stoichiometric coefficients in the balanced chemical equation to determine the relationships between the amounts of reactants and products. For example, the equation tells us that for every 8 moles of H2S that react, 8 moles of H2O are produced and for every 4 moles of O2 that react, 1 mole of S8 is produced.

Step 4: Use the relationships determined in step 3 to fill in the missing values in the table. If you know the amount of one substance, you can use the stoichiometric coefficients to find the amounts of the other substances.

Step 5: Check your work. Make sure that the amounts of reactants and products in the table are consistent with the balanced chemical equation and with each other. If they are not, go back and check your calculations.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a reaction based on balanced chemical equations. Understanding stoichiometry is essential for completing tables that require calculations of moles, mass, or volume of reactants and products.

Balancing Chemical Equations

Balancing chemical equations is the process of ensuring that the number of atoms for each element is the same on both sides of the equation. This is crucial because it reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. A balanced equation provides the correct stoichiometric ratios needed for calculations in stoichiometry.

Gas Laws

Gas laws describe the behavior of gases in relation to pressure, volume, temperature, and the number of moles. Understanding these laws, such as the ideal gas law (PV=nRT), is important when dealing with gaseous reactants and products in a reaction. In the context of the given reaction, gas laws can help in calculating the volumes of gases involved under specific conditions.