Consider the following chemical reaction: 3I2 (s) + 6OH? (aq) 5I? (aq) + IO3? (a

Question

Consider the following chemical reaction: 3I2 (s) + 6OH? (aq) 5I? (aq) + IO3? (aq) + 3H2O(l) a) Calculate TRIANGLE G degree for this reaction as written at25 degree C. (You will also need TRIANGLE G degree f for IO3? (aq) =?128.0 kJ/mol.) b) Is this reaction spontaneous or non-spontaneous at standardstates? c) Calculate the equilibrium constant K for this reaction at25 degree C. d) Given [I?] = 0.10 M and [IO3?] = 0.50 M, determinethe pH required for this reaction to be at equilibrium at25 degree C. Note: For I2 (s) TRIANGLE H degree = 0 kJ/mol TRIANGLE S degree = 116.7 J/K*mol TRIANGLE G degree = 0 kJ/mol Note: For I-(aq) TRIANGLE H degree = 55.9 kJ/mol TRIANGLE S degree = 109.37 J/K*mol TRIANGLE G degree = -51.57 kJ/mol Note: For OH- TRIANGLE H degree = -229.94 kJ/mol TRIANGLE S degree = -10.50 J/K*mol TRIANGLE G degree = -157.30 kJ/mol Note: For H20(l) TRIANGLE H degree = -285.8 kJ/mol TRIANGLE S degree = 69.9 J/K*mol TRIANGLE G degree = -237.2 kJ/mol

Explanation / Answer

b. The reaction will be spontaneous in the delta G The Leader toldyou how to calculate was negative. It is not spontaneous if thedelta G is positive. c. Delta G = -RTlnK    K = e^(- Delta G/(RT))     Temperature in Kelvin and R = 8.314 Delta G in j/mol d. Qc=Kc So, Qc=[.10]^5 * [.50]/[OH-]^6 Set Kc equal to Qc and Solve for the concentration of [OH-] then, pOH= - log([OH-]) pH=14-pOH Hope this helps!

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