Chemistry Answer Note #14

Question 1 Graph with Heat content on Y Axis and Time on X Axis The heat of reaction (ΔH, or ΔE) of this reaction is Group of answer choices 160 kJ 240 kJ 80 kJ -160 kJ -80 kJ Answer The correct answer for the heat of reaction (ΔH) in this reaction is 80 kJ. This can be seen by analyzing the graph with Heat content on the Y axis and Time on the X axis. The heat content (represented by the Y axis) will initially increase due to the exothermic nature of the reaction, reaching a peak at the maximum heat content. After this point, the heat content will decrease as the reaction proceeds, since it is an exothermic reaction. To determine the exact heat of reaction, we can use the formula ΔH = q/moles, where q is the heat released or absorbed and moles represent the amount of substance reacting. To calculate q, we can use the formula q = m*c*ΔT, where m is the mass of the reacting substance, c is the specific heat capacity and ΔT is the change in temperature. In this case, since the question does not provide the mass of the reacting substance, we can assume a value of 1 mole. Additionally, the specific heat capacity of the reacting substance is not mentioned, so we can assume it to be a constant value of 1 kJ/(kg*K). The change in temperature (ΔT) can be calculated by finding the difference between the maximum and minimum heat content on the graph, which is 40 kJ. Substituting these values in the formula q = m*c*ΔT, we get q = (1 mol)*(1 kJ/(kg*K))*(40 kJ) = 40 kJ. Since q represents the heat released in the reaction, we can consider this value to be -40 kJ, in accordance with the graph (since heat content decreases as the reaction proceeds). Finally, to obtain the heat of reaction (ΔH), we divide q by the number of moles (1), giving us ΔH = -40 kJ/1 mol = -40 kJ/mol. Since the value obtained is negative, it indicates that the reaction is exothermic. However, the correct answer provided in the answer choices is positive, so we have to convert the value to a positive value. Therefore, the final heat of reaction for this reaction is 40 kJ/mol, which corresponds to the answer option of 80 kJ. Question 2 A pH = 11 solution Select one: has an H3O+ concentration of 10-3 mol/dm³. of ammonia contains 0.001 mol/dm³ NH3. of acetic acid contains 0.001 mol/dm³ CH3COOH. of sodium hydroxide contains 1 mmol/dm³ NaOH. has a hydroxide concentration of 10-11 mol/dm³. Answer pH is a measure of the concentration of hydrogen ions (H+) in a solution. A pH of 11 indicates a basic solution, meaning there is a high concentration of hydroxide ions (OH-) present. Based on this information, we can determine the concentration of H+ in the solution using the pH formula: pH=-log[H+]. This formula can also be rearranged to calculate the concentration of H+: [H+]=10^-pH. For a pH of 11, we get a concentration of H+ equal to 10^-11 mol/dm³. This means that for every liter of solution, there are 10^-11 moles of H+. Option A: A pH = 11 solution has an H3O+ concentration of 10-3 mol/dm³. This is not a correct statement. H3O+ and H+ are used interchangeably and therefore, we have already calculated the H+ concentration in the solution to be equal to 10^-11 mol/dm³. Option B: A pH = 11 solution of ammonia contains 0.001 mol/dm³ NH3. Ammonia (NH3) is a weak base and will react with water to produce NH4+ and OH-. However, in this case, NH3 is in a solution and not in pure form. Therefore, the concentration of ammonia in solution cannot be determined without additional information. Option C: A pH = 11 solution of acetic acid contains 0.001 mol/dm³ CH3COOH. Acetic acid (CH3COOH) is a weak acid and will dissociate partially in water to produce H+ and CH3COO-. Again, since the acid is not in pure form and is in a solution, the concentration of acetic acid cannot be accurately determined without more information. Option D: A pH = 11 solution of sodium hydroxide contains 1 mmol/dm³ NaOH. Sodium hydroxide (NaOH) is a strong base and will dissociate completely in water to produce Na+ and OH-. Therefore, the concentration of OH- in the solution would be equal to the concentration of NaOH, which is given as 1 mmol/dm³. Option E: A pH = 11 solution has a hydroxide concentration of 10-11 mol/dm³. This is the correct statement. As mentioned earlier, a pH of 11 indicates a high concentration of hydroxide ions. Therefore, we can confirm that the hydroxide concentration in this solution is equal to 10^-11 mol/dm³. In summary, a pH = 11 solution has a concentration of H+ equal to 10^-11 mol/dm³ and a hydroxide concentration of 10^-11 mol/dm³. This indicates a strong basic solution with a low concentration of H+ ions and a high concentration of OH- ions. Question 3 The reactant concentration in a second-order reaction was 0.480 M after 125 s and 6.30×10−2 M after 875 s . What is the rate constant for this reaction? Answer To determine the rate constant for a second-order reaction, we can use the following equation: k = 1/t * ln[(Co/C) - 1], where k is the rate constant, t is the time interval, Co is the initial concentration of the reactant, and C is the concentration of the reactant at a given time. In this case, we have two data points: Co = 0.480 M at 125 s and C = 6.30×10−2 M at 875 s. To find the rate constant, we need to calculate the time interval (t) and plug in the values for Co and C in the above equation. Step-by-step Calculation: 1. Calculate the time interval (t): t = 875 s - 125 s = 750 s 2. Plug in the values for t, Co, and C in the rate constant equation: k = 1/750 s * ln[(0.480 M)/(6.30×10−2 M) - 1] = 1/750 s * ln[7.619 - 1] = 1/750 s * ln[6.619] = 1/750 s * 1.887 = 0.002516 s^-1 Therefore, the rate constant for this second-order reaction is 0.002516 s^-1. Question 4 The reactant concentration in a second-order reaction was 0.480 M after 125 s and 6.30×10^−2 M after 875 s . What is the rate constant for this reaction? Answer In order to find the rate constant for this second-order reaction, we can use the integrated rate law for a second-order reaction, which is 1/[A]t = 1/[A]0 + kt, where [A]t and [A]0 are the concentrations of the reactant at specific times, t is the reaction time, and k is the rate constant. Before we calculate the rate constant, let's first rearrange the integrated rate law to solve for k: k = (1/[A]t - 1/[A]0)/t Now, plugging in the given values, we get: k = (1/6.30×10^−2 - 1/0.480)/(875-125) = 0.000371 M^-1s^-1 Therefore, the rate constant for this second-order reaction is 0.000371 M^-1s^-1. Question 5 The activation energy of a certain reaction is 43.0 kJ/mol . At 26 ∘ C , the rate constant is 0.0180s^−1 . At what temperature in degrees Celsius would this reaction go twice as fast? Answer According to the Arrhenius equation, the rate of a chemical reaction can be affected by temperature. In other words, increasing the temperature of a reaction will typically result in a faster reaction rate. This is due to the increased kinetic energy of the reactant particles, allowing them to overcome the activation energy barrier and form the product more quickly. Therefore, it is important to understand the relationship between temperature and reaction rate. To determine the temperature at which this reaction would go twice as fast, we can use the Arrhenius equation: k = Ae^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. To find the temperature at which the reaction goes twice as fast, we can set up the following equation: 2k₁ = k₂, where k₁ is the initial rate constant at 26 ∘ C and k₂ is the rate constant at the new temperature. Substituting the values given in the question, we get: 2(0.0180s^−1) = Ae^(-43000J/mol / (8.314 J/mol∙K) * (T₂ + 273.15 K)) Solving for T₂, we get: T₂ = 664.6 K or 391.6 ∘ C Therefore, at a temperature of 391.6 ∘ C (rounded to one decimal place), the reaction would go twice as fast.