Consider the hypothetical reaction A-2B for which G98-5.3 kJ/mol are [A] = 7.00x

Question

Consider the hypothetical reaction A-2B for which G98-5.3 kJ/mol are [A] = 7.00x10-M and [B] = 1.300 M at a temperature of 298 K. Calculate G for the reaction at these concentrations. a) 8.3 kJ/molb) 7.6 kJ/molc) 0.60 kJ/mol d) 1.4 x 10 kJ/mol e) none of these Based on your answer to Problem 2, will the reaction shift to the right (toward products) to reach equilibrium? a) Yes 2. In a certain experiment, the initial concentrations 4 3. b) No c) impossible to determine 4. Use the given information to calculate the equilibrium constant, K, for the following equilibrium reaction at 25°C. 2 H20 (g)2 H2 (g) + O2 (g) GF[H20(g)]--228.6 kJ/mol GfTH20()] =-237.2 kJ/mol a) 1.4×1080 b) 1.8×102 c) 5.2 d) 7.2×10-81 e) 0.83

Explanation / Answer

G = G° + RT*ln(Q)

G = -5300 + 8.314*298*ln((1.3^2)/(7*10^-3))

G = 8293.38 = 8.3 kJ

choose A

3.

the reaction ahs a positive G; therefore it favours reactants

4.

G = Gprod - Greact = 2*0 +0 - 2*-228.6 =457.2 kJ/mol

G = -RT*lnK

K = exp(457.2/(-8.314*298) = 0.83149

nearest is E

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