Chemistry Answer Note #15

Question 1 Clearly explain in detail what makes H2O a polar molecule and why H-bonds can form between H2O molecules. Answer H2O, also known as water, is a molecule composed of two hydrogen atoms and one oxygen atom. It is commonly referred to as a polar molecule due to its unique structure and electronegativity differences between the atoms. The shape of the H2O molecule is bent, with the two hydrogen atoms bonded to the oxygen atom at an angle of approximately 105 degrees. This bent shape is caused by the unequal sharing of electrons between the atoms. Oxygen has a higher electronegativity than hydrogen, meaning it has a greater attraction for electrons. As a result, the shared electrons in the covalent bonds between the atoms spend more time towards the oxygen atom, creating a partial negative charge (δ-) on the oxygen and a partial positive charge (δ+) on the hydrogen atoms. This unequal distribution of charge within the molecule causes H2O to have a dipole or a separation of positive and negative charges. This dipole moment is what makes H2O a polar molecule. The δ- and δ+ charges create an electric field within the molecule, making it attracted to other polar molecules and ions. Now, let's explore why H-bonds can form between H2O molecules. Hydrogen bonds (Hbonds) are a type of intermolecular forces that occur between polar molecules or polar covalent bonds within a molecule. In the case of H2O, the δ- and δ+ charges of the molecule can interact with each other, forming H-bonds. H-bonds occur between the oxygen atom of one H2O molecule and the hydrogen atom of another H2O molecule. The δ+ hydrogen atom is attracted to the δ- oxygen atom of another molecule, creating a weak electrostatic bond. This bond is about 5-10% as strong as a covalent bond, but it plays a vital role in the unique properties of water. One of the most important properties of H-bonds is their ability to create surface tension and cohesion within water. The H-bonds hold the water molecules together, allowing water to resist external forces and form droplets. This cohesive force also allows water molecules to stick together, creating surface tension which allows insects and small organisms to walk on the surface of water. In summary, H2O is a polar molecule due to its bent shape and unequal sharing of electrons between atoms. The polarity of the molecule allows for the formation of H- bonds between water molecules, resulting in unique properties such as surface tension and cohesion. H-bonds are crucial for the existence of life on Earth, making water an essential molecule for all living organisms. Question 2 Determine the pH of an HNO2 solution of each concentration. In which cases can you not make the simplifying assumption that x is small? a. 0.500 M b. 0.100 M c. 0.0100 M Answer The pH of an HNO2 solution can be determined by utilizing the acid dissociation constant, Ka, of nitrous acid (HNO2). The higher the concentration of the solution, the lower the pH will be, as more HNO2 molecules will dissociate and increase the concentration of H+. In order to make the simplifying assumption that x is small, the initial concentration of HNO2 must be small enough that the change in concentration due to dissociation is negligible. a. For a 0.500 M HNO2 solution, the pH can be determined by using the Ka value of 4.5x10^-4 and the dissociation reaction: HNO2 + H2O ⇌ H3O+ + NO2-. We can set up a simple ICE (initial change equilibrium) table to calculate the concentration of H+, which can then be used to find the pH. Initial: [HNO2] = 0.500 M, [H3O+] = 0, [NO2-] = 0 Change: -x, +x, +x Equilibrium: [HNO2] = 0.500 - x, [H3O+] = x, [NO2-] = x The Ka expression for this reaction is: Ka = [H3O+][NO2-]/[HNO2] = (x)(x)/(0.500 - x) = 4.5x10^-4, which simplifies to x^2 = 4.5x10^-4(0.500 - x). This quadratic equation can be solved using the quadratic formula, resulting in two solutions: x = 0.0115 M and x = 0.0050 M. Since x represents the concentration of H+, we can eliminate the negative solution and use the positive solution to calculate the pH: pH = -log(0.0115) = 1.94. b. For a 0.100 M HNO2 solution, we can follow the same procedure as in part a to determine the pH. However, since the initial concentration of HNO2 is smaller, the resulting quadratic equation will only have one positive solution: x = 0.0037 M. Thus, the pH of a 0.100 M HNO2 solution is pH = -log(0.0037) = 2.43. In this case, the simplifying assumption can be made because the change in concentration due to dissociation is small enough to be negligible. c. For a more dilute solution of 0.0100 M HNO2, the simplifying assumption cannot be made. Following the same procedure as in part a, we get a quadratic equation with no positive solutions. This indicates that the change in concentration due to dissociation is not negligible and the simplifying assumption cannot be made. In this case, a more accurate method of calculation, such as the quadratic formula or an iterative method, can be used to determine the pH. The resulting pH value will be slightly different from the one that would have been calculated using the simplifying assumption. Question 3 What Reaction goes to 100% Yield Write a balanced equation List the 2 most important reasons reaction goes to 100% completion Hint. product is one of the most abundant materials states of matter change Answer As a high-achieving college student, I understand the importance of achieving a 100% yield in a chemical reaction. A 100% yield means that all of the reactants have been converted into products, indicating a successful and efficient reaction. In this response, I will discuss the factors that contribute to a reaction going to 100% completion and provide a balanced equation to showcase this concept. Firstly, it is crucial to understand that a reaction reaching 100% yield is not always feasible in the real world. This is because there are many factors that can affect the yield of a reaction, such as impurities, side reactions, and incomplete reactions. However, there are two main reasons that can contribute to a reaction reaching 100% completion, which are an excess of reactants and favorable reaction conditions. An excess of reactants is one of the most important reasons for a reaction to reach 100% yield. This means that the reactants are present in a higher amount than what is needed for the reaction to occur. When there is an excess of reactants, it ensures that all of the available reactants are utilized, leaving no room for incomplete reactions. Additionally, an excess of reactants also helps to drive the reaction forward and promote the formation of products. The second important reason for a reaction going to 100% completion is favorable reaction conditions. This includes factors such as temperature, pressure, and catalysts. These conditions can influence the rate of the reaction and help to overcome any energy barriers, allowing the reaction to occur more efficiently. For example, a higher temperature can increase the kinetic energy of the particles, leading to more frequent and successful collisions between reactant molecules. To showcase the concept of a reaction going to 100% yield, let us consider the balanced equation for the formation of water from hydrogen and oxygen gas: 2H2 + O2 --> 2H2O In this reaction, we can see that for every two molecules of hydrogen and one molecule of oxygen, we will get two molecules of water. In theory, this reaction can reach 100% yield if we have an excess of both hydrogen and oxygen gas and favorable reaction conditions. In terms of academic sources, I have consulted the following resources to provide an accurate and comprehensive answer: In conclusion, while it is not always possible to achieve a 100% yield in a chemical reaction, an excess of reactants and favorable reaction conditions are the two most important factors that can contribute to this outcome. By understanding these concepts and utilizing them in our experiments, we can increase the chances of reaching a 100% yield and obtaining the desired products. Question 4 Question: “The reactant concentration in a first-order reaction was 5.00×10^−2 M after 45.0 s and 1.20×10^−3 M after 80.0 s. What is the rate constant for this reaction?” Response: To find the rate constant for this first-order reaction, we can use the integrated rate law equation for a first-order reaction: ln[A]t = -kt + ln[A]0 Where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and ln is the natural logarithm function. By plugging in the given values and solving for k, we get: ln(1.20×10^-3 M) = -k(80.0 s) + ln(5.00×10^-2 M) -7.123 = 80.0k + -2.996 -80.0k = -4.127 k = 5.158 × 10^-2 s^-1 Therefore, the rate constant for this reaction is 5.158 × 10^-2 s^-1. To further understand the concept of first-order reactions and rate constants, here are some useful sources: 1. “First-Order Reactions.” Chemistry LibreTexts, Libretexts, 20 Aug. 2020, chem.libretexts.org/Courses/North_Central_College/Chemistry_211_Chemical_Kinetics/ 05_Concentration_and_Rate_Calculations/5.03_First_Order_Reactions. This source provides a comprehensive explanation of first-order reactions, including the definition, equations, and calculations. 2. Chemistry, The Central Science: Brown, Theodore L., et al. “Chapter 14: Chemical Kinetics.” Pearson, 2017. This textbook chapter discusses the concepts of reaction rates and rate laws, including first-order reactions and rate constants. It also provides example problems and step-by-step solutions. 3. “Rate Laws.” Khan Academy, Khan Academy, 3 Apr. 2019, www.khanacademy.org/science/chemistry/chem-kinetics/rate-law/a/the-rate-law. This online resource includes a video explanation of rate laws and their equations, as well as practice problems for further understanding. With a combination of these sources, you can gain a thorough understanding of first-order reactions and rate constants, and effectively solve problems like the one given in this question. Remember to always check your calculations and use units for accuracy. Good luck! Answer To find the rate constant for this first-order reaction, we can use the integrated rate law equation for a first-order reaction: ln[A]t = -kt + ln[A]0 Where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and ln is the natural logarithm function. By plugging in the given values and solving for k, we get: ln(1.20×10^-3 M) = -k(80.0 s) + ln(5.00×10^-2 M) -7.123 = -80.0k + -2.996 -80.0k = -4.127 k = 5.158 × 10^-2 s^-1 Therefore, the rate constant for this reaction is 5.158 × 10^-2 s^-1. Step-by-Step Calculation: 1. Start with the integrated rate law equation: ln[A]t = -kt + ln[A]0 2. Plug in the given values for [A]t (1.20×10^-3 M) and [A]0 (5.00×10^-2 M). 3. Rearrange the equation to solve for k: -kt = ln[A]t - ln[A]0 4. Substitute the values for [A]t and [A]0: -kt = ln(1.20×10^-3 M) - ln(5.00×10^-2 M) 5. Use a calculator to solve for the natural logarithm functions: ln(1.20×10^-3 M) = 7.123 and ln(5.00×10^-2 M) = -2.996 6. Substitute these values into the equation: -kt = -7.123 - (-2.996) 7. Simplify: -kt = -4.127 8. Divide both sides by -80.0 s: k = 5.158 × 10^-2 s^-1 9. Use proper scientific notation: k = 5.158 × 10^-2 s^-1. Dimensional Analysis: In this problem, dimensional analysis can be used to check the accuracy of the final answer and to convert units if necessary. The given units are in molarity (M) and time (s). Since the rate constant is in units of s^-1, the units match the units of time in the given data. To confirm the accuracy of the calculation, the units can be canceled out using dimensional analysis: 5.158 × 10^-2 M / 80.0 s M (unit of concentration) cancels out with 1/M (unit of rate constant) Leaving us with 5.158 s^-1, the correct unit for the rate constant. In-Depth Explanation: First-order reactions are chemical reactions where the rate of the reaction is directly proportional to the concentration of one reactant. The rate constant (k) is a proportionality constant that relates the reaction rate to the concentration of the reactant. The integrated rate law equation for a first-order reaction, as shown in the response above, can be derived from the rate law: rate = k[A], where [A] is the concentration of the reactant. This equation allows us to calculate the rate constant using the concentration of the reactant at different times. The given data gave us the concentration of the reactant (A) at two different times (t), which we can use to solve for the rate constant (k). By substituting the values into the integrated rate law equation and rearranging, we can solve for k. The natural logarithm function is used in the integrated rate law equation because it allows us to easily isolate the rate constant on one side of the equation. Question 5 The reactant concentration in a zero-order reaction was 8.00×10−2 M after 155s and 3.00×10−2 M after 355s . What is the rate constant for this reaction? Answer In a zero-order reaction, the reaction rate is independent of the concentration of the reactant. This means that the reaction rate remains constant throughout the reaction and is determined by the rate constant, k. To find the rate constant, we can use the equation: [A]t = [A]0 - kt, where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, and k is the rate constant. In this case, we have two data points: [A]0 = 8.00×10−2 M at t=155s and [A]t = 3.00×10−2 M at t=355s. Plugging these values into the equation, we get: 3.00×10−2 M = 8.00×10−2 M - k(355s-155s) Simplifying, we get: k = (8.00×10−2 M - 3.00×10−2 M)/(355s-155s) = 5.00×10−2 M/200s = 2.5×10^-4 M/s. Therefore, the rate constant for this reaction is 2.5×10^-4 M/s.