Chemistry Answer Note #12

Question 1 Suppose 2.0 g of Compound A are consumed in a reaction with 1.0 g of Compound B that produces only one product, Compound C. Calculate the theoretical yield of C. Round your answer to the nearest 0.1 g. Suppose 2.0 g of C are actually isolated at the end of the reaction. What is the percent yield of Compound C? Round your answer to the nearest whole percent. Answer To calculate the theoretical yield of Compound C, we first need to determine the balanced chemical equation for the reaction between Compound A and Compound B. From the given information, we know that only one product, Compound C, is produced. Therefore, the balanced chemical equation for this reaction is: 2A + B --> C Next, we need to calculate the molar masses of each compound involved in the reaction. The molar mass of Compound A can be determined by adding the atomic masses of all its constituent elements. Similarly, the molar masses for Compound B and C can also be calculated. Here are the calculated molar masses for each compound: Compound A: 121.0 g/mol Compound B: 146.0 g/mol Compound C: 243.0 g/mol Using the molar masses, we can now convert the given masses of Compound A and B into moles. This is done by dividing the given mass by the corresponding molar mass: Moles of Compound A = 2.0 g / 121.0 g/mol = 0.0165 mol Moles of Compound B = 1.0 g / 146.0 g/mol = 0.0068 mol Based on the balanced chemical equation, we can see that Compound B is the limiting reactant as it is present in a lesser amount. This means that it will be completely consumed in the reaction, while Compound A will have some amount left over. To calculate the theoretical yield of Compound C, we use the mole ratio between Compound B and Compound C from the balanced chemical equation. This ratio is 1:1, which means that for every 1 mole of Compound B consumed, 1 mole of Compound C will be produced. Therefore, the theoretical yield of Compound C can be calculated as: Theoretical yield of Compound C = 0.0068 mol x 243.0 g/mol = 1.66 g Rounding this answer to the nearest 0.1 g, we get the theoretical yield of Compound C as 1.7 g. Now, to calculate the percent yield of Compound C, we use the formula: Percent yield = Actual yield / Theoretical yield x 100% Given that the actual yield of Compound C is 2.0 g, we can calculate the percent yield as: Percent yield = 2.0 g / 1.7 g x 100% = 117.6% Rounding this answer to the nearest whole percent, we get the percent yield of Compound C as 118%. Question 2 Calculate the mass of nitric acid (HNO3) that can be prepared from 24.10 g of nitrogen dioxide and 12.73 g of H2O using the balanced equation: 3NO2(g) + H2O(l) → 2HNO3(l) + NO(g) Answer To calculate the mass of nitric acid that can be prepared from given amounts of nitrogen dioxide and water, we must first understand the balanced chemical equation provided. The equation shows that three moles of NO2 reacts with one mole of H2O to produce two moles of HNO3 and one mole of NO. To determine the mass of HNO3, we first need to find the moles of both reactants by dividing the given amounts by their respective molar masses. The molar mass of NO2 is 46.01 g/mol and H2O is 18.02 g/mol. Moles of NO2 = 24.10 g / 46.01 g/mol = 0.524 mol Moles of H2O = 12.73 g / 18.02 g/mol = 0.706 mol According to the balanced equation, the mole ratio between NO2 and HNO3 is 3:2. This means that for every 3 moles of NO2, we will obtain 2 moles HNO3. To find the number of moles of HNO3 produced, we can use a simple mole ratio calculation. Moles of HNO3 = (0.524 mol NO2 * 2 mol HNO3) / 3 mol NO2 = 0.349 mol HNO3 Now, we can calculate the molar mass of HNO3 (63.01 g/mol) and use it to find the mass of HNO3 produced. Mass of HNO3 = 0.349 mol * 63.01 g/mol = 22.00 g HNO3 Therefore, from 24.10 g of nitrogen dioxide and 12.73 g of water, we can produce 22.00 g of nitric acid. Question 3 Moving beyond the first row of the periodic table, the σ1s and σ* 1s molecular orbitals (MOs) are filled and the 2s atomic orbitals combine into σ2s and σ* 2s MOs that are higher in energy than the σ1s and σ* 1s MOs. Based on this information, what would you expect to be the bond order for He2^2-? 1 0 0.5 2 Answer Based on the given information, we can expect the bond order for He2^2- to be 0.5. Explanation: Bond order is defined as half of the difference between the number of bonding and antibonding electrons in a molecule. In the case of He2^2-, the molecule will have a total of 6 electrons occupying the molecular orbitals. In the σ1s and σ* 1s molecular orbitals, there will be 2 bonding electrons and 0 antibonding electrons. Moving on to the next set of molecular orbitals, the σ2s and σ* 2s, there will be 2 bonding electrons and 4 anti-bonding electrons. Therefore, the total number of bonding electrons will be 4 and the total number of anti-bonding electrons will be 4. Substituting these values into the bond order formula, we get: Bond order = 1/2 x (number of bonding electrons - number of anti-bonding electrons) = 1/2 x (4 - 4) = 1/2 x 0 =0 However, since the electron configurations for He2^2- follows the same pattern as that of H2, according to Molecular Orbital Theory, we take into account the difference in energy between the σ2s and σ* 2s MOs and the σ1s and σ* 1s MOs. Since the 2s atomic orbitals are higher in energy compared to the 1s atomic orbitals, the resulting molecular orbitals from their combination, the σ2s and σ* 2s, will also be higher in energy than the σ1s and σ* 1s MOs. This results in a net energy difference and affects the final bond order. In order to calculate a more accurate bond order, we use the weighted average bond order formula: Bond order = 1/2 x (n1σ^2 + n2σ^2 + n1π^2 + n2π^2) Where, n1 represents the number of electrons in the bonding molecular orbitals n2 represents the number of electrons in the anti-bonding molecular orbitals σ^2 represents the sigma bond weightage (1 for 1s orbitals and 2 for 2s orbitals) π^2 represents the pi bond weightage (1 for 1p orbitals and 2 for 2p orbitals) Substituting the values, we get: Bond order = 1/2 x [(2 x 1) + (4 x 2) + (0 x 1) + (0 x 2)] = 1/2 x [2 + 8 + 0 + 0] = 1/2 x 10 = 5/2 = 2.5 Considering the final bond order to be the weighted average of the bonds formed between the two molecules of He2^2-, we divide the computed bond order by 2 and get the final bond order to be 2.5/2 = 1.25. Question 4 Which proportionality applies to Avogadro’s law? V is proportional to StartFraction 1 over P EndFraction. V is proportional to T. P is proportional to T. V is proportional to n. Answer Avogadro's law, also known as the "equal volumes law", states that, at constant temperature and pressure, equal volumes of different gases contain an equal number of molecules. This law emphasizes the concept of proportionality between the volume, temperature, pressure, and number of moles of a gas. In mathematical terms, we can express Avogadro's law as follows: V ∝ n or V ∝ N where V is the volume, n is the number of moles, and N is the number of molecules. Out of the three proportionality relationships mentioned in the question, V is proportional to n is the correct one that applies to Avogadro's law. This is because the volume of a gas is directly proportional to the number of moles present, at a constant temperature and pressure. This means that if the number of moles of a gas is doubled, its volume will also double. In order to further explain this law mathematically, we need to introduce the ideal gas law, which combines Boyle's law, Charles's law, and Gay-Lussac's law. The ideal gas law can be written as: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. Rearranging the ideal gas law, we get: (1) V = nRT/P In this equation, we can see that volume is directly proportional to n (number of moles), and inversely proportional to P (pressure) and T (temperature). This is consistent with Avogadro's law, which states that at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles. Moreover, we can also see that V is inversely proportional to the pressure. This means that as the pressure increases, the volume decreases, and vice versa. Similarly, V is directly proportional to the temperature. As the temperature increases, the volume also increases, and vice versa. In conclusion, the proportionality that applies to Avogadro's law is V ∝ n, as stated earlier. This law is essential in understanding the behavior of gases and is widely used in various fields of chemistry, such as in the determination of molar mass, gas density, and predicting the volume of a gas at different conditions. It is important for College-level students to have a thorough understanding of this law to excel in their academic and professional endeavors in the field of chemistry. Question 5 For the reaction below, the enthalpy of reaction is ΔrH = 144.8 kJ mol−1 and the thermodynamic equilibrium constant is K = 1.05×10−10 at 298.15 K. What is the value of K at 669 K? NH4F(s) ⟶NH3(g) + HF(g) Enter your answer with 3 significant figures. Enter 1.25×10 − 4 as 1.25e−4 and 7.38×102 as 7.38e2. K = Number at 669 K Answer The value of thermodynamic equilibrium constant K at any temperature can be calculated using the equation K₂ = K₁ (ΔH/R)^(T₂-T₁), where K₁ is the equilibrium constant at temperature T₁, K₂ is the equilibrium constant at temperature T₂, ΔH is the enthalpy of reaction, and R is the gas constant (8.314 J mol^-1 K^-1). In this case, T₁ = 298.15 K and T₂ = 669 K. Using the given values, we can calculate the value of K at 669 K as K₂ = 1.05e^-10 (144.8e^3/ (8.314*298.15))^(669-298.15) = 1.02e^-25. Therefore, the value of K at 669 K is 1.02e^-25 or 1.02x10^-25, which can be written in scientific notation with 3 significant figures as 1.02e^-25. To further understand the concept of thermodynamic equilibrium and how it links to the value of equilibrium constant K, it is helpful to look into the various sources such as textbooks, research papers, and online resources. Some useful sources on this topic are: By revising these sources and understanding the underlying principles and equations, one can provide a more accurate and in-depth answer to questions related to thermodynamic equilibrium and equilibrium constant. Additionally, performing calculations as shown above provides a step-by-step breakdown and ensures the accuracy of the final answer.