14) Which of the following statement(s) IV are correct for the reaction below. Pt,H_(2)(580torr)vert HCl(0.0016M)Vert Pd^wedge 200M(s) (E^circ Pa^ast /Pd=0.987V)(760torr:1atm) The reaction type is spontaneous II. Ecell=0.809V III- Keq=2.19times 10^33 A) Only I B) I and II C I and III D) II and III

Question

14) Which of the following statement(s) IV are correct for the reaction below. Pt,H_(2)(580torr)vert HCl(0.0016M)Vert Pd^wedge 200M(s) (E^circ Pa^ast /Pd=0.987V)(760torr:1atm) The reaction type is spontaneous II. Ecell=0.809V III- Keq=2.19times 10^33 A) Only I B) I and II C I and III D) II and III

Answer 1

To determine which statements are correct for the given reaction, we need to analyze each statement individually.Statement I: The reaction type is spontaneous.To determine if the reaction is spontaneous, we need to calculate the cell potential (E_cell) and compare it to the standard cell potential (E^°_cell). If E_cell is positive, the reaction is spontaneous.Statement II: E_cell = 0.809 V.To calculate E_cell, we can use the Nernst equation:E_cell = E^°_cell - (0.0592/n) * log(Q)where n is the number of moles of electrons transferred in the reaction, and Q is the reaction quotient.Statement III: Keq = 2.19 x 10^33.To calculate the equilibrium constant (K_eq), we can use the relationship between E_cell and K_eq:E^°_cell = (0.0592/n) * log(K_eq)Now let's analyze each statement:Statement I: To determine if the reaction is spontaneous, we need to calculate E_cell. We can use the given values to calculate E_cell:E^°_cell = 0.987 Vn = 2 (since 2 moles of electrons are transferred in the reaction)P_H2 = 580 torr = 580/760 atm = 0.7639 atmP_HCl = 0.0016 M * 0.987 V = 0.0015768 atmUsing the Nernst equation:E_cell = 0.987 - (0.0592/2) * log((0.7639/760) * (0.0015768/0.0016))E_cell = 0.987 - 0.0296 * log(0.00100658)E_cell = 0.987 - 0.0296 * (-3.007)E_cell = 0.987 + 0.0894E_cell = 1.0764 VSince E_cell is positive, the reaction is spontaneous. Therefore, statement I is correct.Statement II: E_cell = 0.809 V.From our calculation above, we found that E_cell = 1.0764 V, which is not equal to 0.809 V. Therefore, statement II is incorrect.Statement III: Keq = 2.19 x 10^33.To calculate K_eq, we can use the relationship between E^°_cell and K_eq:E^°_cell = (0.0592/n) * log(K_eq)0.987 = (0.0592/2) * log(K_eq)0.987 = 0.0296 * log(K_eq)log(K_eq) = 33.38K_eq = 10^33.38 = 2.19 x 10^33Therefore, statement III is correct.Based on our analysis, the correct answer is C) I and III.

Soru

14) Which of the following statement(s) IV are correct for the reaction below. Pt,H_(2)(580torr)vert HCl(0.0016M)Vert Pd^wedge 200M(s) (E^circ Pa^ast /Pd=0.987V)(760torr:1atm) The reaction type is spontaneous II. Ecell=0.809V III- Keq=2.19times 10^33 A) Only I B) I and II C I and III D) II and III

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