Let's solve the problem step by step as requested. --- ## **Given:** - **Reaction:** \( \text{A(g)} + \text{B(g)} \rightleftharpoons 2\text{C(g)} \) - **Equilibrium constant,** \( K = 4.66 \times 10^{-3} \) - **Initial moles:** \( [\text{A}]_ = .300 \) mol, \( [\text{B}]_ = .100 \) mol, \( [\text{C}]_ = \) mol - **Volume:** \( 1 \) L, so molarity = moles --- ## **Step 1: Write the equilibrium expressions** Let \( x \) be the change in concentration (mol/L) of A and B that react: | | A | B | C | |-----|-------------|-------------|-------------| | I | .300 | .100 | | | C | -x | -x | +2x | | E | .300 - x | .100 - x | 2x | At equilibrium: - \( [\text{A}] = .300 - x \) - \( [\text{B}] = .100 - x \) - \( [\text{C}] = 2x \) The equilibrium constant expression is: \[ K = \frac{[\text{C}]^2}{[\text{A}][\text{B}]} \] Substitute the equilibrium concentrations: \[ 4.66 \times 10^{-3} = \frac{(2x)^2}{(.300-x)(.100-x)} \] --- ## **Step 2: Write the quadratic equation for \( x \)** \[ 4.66 \times 10^{-3} = \frac{4x^2}{(.300-x)(.100-x)} \] Multiply both sides by \( (.300-x)(.100-x) \): \[ 4.66 \times 10^{-3} (.300-x)(.100-x) = 4x^2 \] Expand \( (.300-x)(.100-x) \): \[ (.300-x)(.100-x) = .030 - .300x - .100x + x^2 = .030 - .400x + x^2 \] So, \[ 4.66 \times 10^{-3} (.030 - .400x + x^2) = 4x^2 \] Expand: \[ 4.66 \times 10^{-3} \times .030 - 4.66 \times 10^{-3} \times .400x + 4.66 \times 10^{-3} x^2 = 4x^2 \] \[ 1.398 \times 10^{-4} - 1.864 \times 10^{-3} x + 4.66 \times 10^{-3} x^2 = 4x^2 \] Bring all terms to one side: \[ 1.398 \times 10^{-4} - 1.864 \times 10^{-3} x + 4.66 \times 10^{-3} x^2 - 4x^2 = \] \[ 1.398 \times 10^{-4} - 1.864 \times 10^{-3} x - 3.99534 x^2 = \] Or, for simplicity, multiply both sides by \( 10^4 \) to clear decimals: \[ 1.398 - 18.64 x - 39953.4 x^2 = \] But let's keep it in standard quadratic form for clarity: \[ -3.99534 x^2 - 1.864 \times 10^{-3} x + 1.398 \times 10^{-4} = \] Or to make coefficients positive, multiply both sides by \(-1\): \[ 3.99534 x^2 + 1.864 \times 10^{-3} x - 1.398 \times 10^{-4} = \] --- ## **Step 3: Solve the quadratic equation** The quadratic equation is: \[ 3.99534 x^2 + 1.864 \times 10^{-3} x - 1.398 \times 10^{-4} = \] This is of the form \( ax^2 + bx + c = \): - \( a = 3.99534 \) - \( b = 1.864 \times 10^{-3} \) - \( c = -1.398 \times 10^{-4} \) Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Calculate the discriminant: \[ b^2 - 4ac = (1.864 \times 10^{-3})^2 - 4 \times 3.99534 \times (-1.398 \times 10^{-4}) \] \[ = 3.475 \times 10^{-6} + 2.232 \times 10^{-3} \] \[ = .0022355 \] Now plug in values: \[ x = \frac{-(1.864 \times 10^{-3}) \pm \sqrt{.0022355}}{2 \times 3.99534} \] \[ \sqrt{.0022355} \approx .04733 \] \[ x = \frac{-.001864 \pm .04733}{7.99068} \] Take the positive root (since x cannot be negative): \[ x = \frac{-.001864 + .04733}{7.99068} \] \[ x = \frac{.045466}{7.99068} \] \[ x \approx .00569 \] --- ## **Step 4: Find equilibrium concentrations** - \( [\text{A}] = .300 - x = .300 - .00569 = .294 \) M - \( [\text{B}] = .100 - x = .100 - .00569 = .0943 \) M - \( [\text{C}] = 2x = 2 \times .00569 = .0114 \) M --- ## **### Final Answers** ### (a) **Quadratic equation:** \[ 3.99534 x^2 + 1.864 \times 10^{-3} x - 1.398 \times 10^{-4} = \] ### (b) **Value of \( x \):** \[ x \approx .00569 \] ### (c) **Equilibrium concentrations:** - \( [\text{A}]_{eq} = .294 \) M - \( [\text{B}]_{eq} = .0943 \) M - \( [\text{C}]_{eq} = .0114 \) M --- **Let me know if you need the quadratic completed by the square!**

Let's find the number of choices that need to be eliminated for guessing to be advantageous. --- ## **Given:** - **Points for a correct answer:** \( 7 \) - **Points for an incorrect answer:** \( 0 \) - **Points for a blank answer:** \( 2 \) - **Total choices:** \( 5 \) --- ## **Expected Value Calculation** The expected value \( E \) of guessing can be calculated as follows: 1. **Probability of Correct Answer (\( P_C \)):** \[ P_C = \frac{1}{n} \] where \( n \) is the number of choices remaining. 2. **Probability of Incorrect Answer (\( P_I \)):** \[ P_I = \frac{n-1}{n} \] 3. **Expected Value \( E \):** \[ E = P_C \times \text{points for correct} + P_I \times \text{points for incorrect} + P_B \times \text{points for blank} \] where \( P_B \) is the probability of leaving the answer blank, which can be assumed to be the remaining probability when guesses are made. Assuming you don't guess, you would score \( 2 \) points on average if you leave it blank. ### **Expected Value Formula:** Thus, the expected value can be simplified to: \[ E = \left( \frac{1}{n} \times 7 \right) + \left( \frac{n-1}{n} \times 0 \right) + \left( \frac{1}{n} \times 2 \right) \] This can be rewritten as: \[ E = \frac{7}{n} + \frac{2}{n} = \frac{7 + 2}{n} = \frac{9}{n} \] ### **Set Up the Inequality:** We want this expected value to be greater than \( 2 \): \[ \frac{9}{n} > 2 \] ### **Solving the Inequality:** Multiply both sides by \( n \) (assuming \( n > 0 \)): \[ 9 > 2n \] Divide by \( 2 \): \[ n < \frac{9}{2} = 4.5 \] Since \( n \) must be a whole number, we can conclude: \[ n \leq 4 \] --- ## **Determine Choices Eliminated:** Initially, there are \( 5 \) choices. To make \( n \) equal to \( 4 \) or less: - If \( n = 4 \), then \( \text{choices eliminated} = 5 - 4 = 1 \) - If \( n = 3 \), then \( \text{choices eliminated} = 5 - 3 = 2 \) - If \( n = 2 \), then \( \text{choices eliminated} = 5 - 2 = 3 \) - If \( n = 1 \), then \( \text{choices eliminated} = 5 - 1 = 4 \) Thus, to make guessing advantageous, you must eliminate at least **1 choice**. --- ## **### Final Answer** You must eliminate **1 choice** for guessing to be advantageous (expected value > 2).

# Formation of Frenkel Defects Frenkel defects occur when an ion is displaced from its lattice site and occupies an interstitial site, resulting in a vacancy at the original site. The formation of Frenkel defects can be represented by the following reactions: ## **(a) Formation of Frenkel Defect in \( \text{MO} \)** For a metal oxide \( \text{MO} \), where \( \text{M} \) is a metal cation and \( \text{O} \) is an anion, the formation of a Frenkel defect can be represented as: \[ \text{M}^{2+}_{\text{lattice}} + \text{O}^{2-}_{\text{lattice}} \rightleftharpoons \text{M}^{2+}_{\text{interstitial}} + \text{V}_{\text{M}} + \text{V}_{\text{O}} \] Where: - \( \text{M}^{2+}_{\text{lattice}} \) is the metal ion in the lattice, - \( \text{O}^{2-}_{\text{lattice}} \) is the oxide ion in the lattice, - \( \text{M}^{2+}_{\text{interstitial}} \) is the metal ion that has moved to an interstitial site, - \( \text{V}_{\text{M}} \) is a vacancy where the metal ion was located, - \( \text{V}_{\text{O}} \) is the vacancy created in the oxide lattice. ## **(b) Formation of Frenkel Defect in \( \text{CaF}_2 \)** For calcium fluoride \( \text{CaF}_2 \), where \( \text{Ca}^{2+} \) is the cation and \( \text{F}^- \) is the anion, the formation of a Frenkel defect can be represented as: \[ \text{Ca}^{2+}_{\text{lattice}} + 2\text{F}^-_{\text{lattice}} \rightleftharpoons \text{Ca}^{2+}_{\text{interstitial}} + \text{V}_{\text{Ca}} + 2\text{V}_{\text{F}} \] Where: - \( \text{Ca}^{2+}_{\text{lattice}} \) is the calcium ion in the lattice, - \( \text{F}^-_{\text{lattice}} \) is the fluoride ion in the lattice, - \( \text{Ca}^{2+}_{\text{interstitial}} \) is the calcium ion that has moved to an interstitial site, - \( \text{V}_{\text{Ca}} \) is a vacancy where the calcium ion was located, - \( \text{V}_{\text{F}} \) is the vacancy created in the fluoride lattice. --- ### **Summary** - **MO Frenkel Defect Reaction:** \[ \text{M}^{2+}_{\text{lattice}} + \text{O}^{2-}_{\text{lattice}} \rightleftharpoons \text{M}^{2+}_{\text{interstitial}} + \text{V}_{\text{M}} + \text{V}_{\text{O}} \] - **CaF₂ Frenkel Defect Reaction:** \[ \text{Ca}^{2+}_{\text{lattice}} + 2\text{F}^-_{\text{lattice}} \rightleftharpoons \text{Ca}^{2+}_{\text{interstitial}} + \text{V}_{\text{Ca}} + 2\text{V}_{\text{F}} \]