Which of the following reactions is unfavorable at low temperature but becomes f

Question

Which of the following reactions is unfavorable at low temperature but becomes favorable as the temperature increases? 2 CO(g) + O_2(g) rbarr CO_2(g); Delta H degree = -566 kJ; Delta S degree = -173 J/K a. 2 H_2O(g) rbarr 2 H_2(g); Delta H degree = 484 kJ; Delta S degree = 149 J/K c. PbCl_2(s) rbarr Pb^2+ (aq) + O_2(g); Delta H degree = 23.4 kJ; Delta S degree = -12.5 J/K The following reaction is spontaneous as written; Zn(s) + Cu^2+ (aq) rbarr Zn^2+ (aq) + Cu(s) When a reaction is at equilibrium, which of the following statements is TRUE? Delta G = Delta G degree a. In K_eq = 0 b. Delta G degree = 0 c. Q = 0 d. Delta G = 0 Consider the gas-phase reaction of hydrogen atom, H_2, and oxygen, O_2 to 2 H_2(g) + O_2(g) rightarrow 2 H_2 O(g) and the following thermodynamic data: Calculate the value of the equilibrium constraint for this reaction under standard conditions. For a certain process, Delta S (system) > 0 and Delta S(surroundings) > 0. The process: is spontaneous. A. is exothermic. B. decreases process at 500 K, Delta G = -3.0 kJ and Delta H = -23.0 kJ. If the process is carried out reversibly, calculate the amount (in kJ) of useful work that can be performed. A certain chemical reaction out at constant temperature and pressure has Delta S > 0. What can you conclude about the spontaneity of this reaction? The reaction is not spontaneous if it is endothermic and the temperature is sufficiently high. a. The reaction is not spontaneous at any temperature.

Explanation / Answer

(17)

(b) 2 H2O (g) --------> 2 H2 (g) + O2 (g)

deltaG = deltaH - TdeltaS

for reaction to be spontaneous, deltaG = -ve,

Since deltaH is given positive and deltaS is given positive, to be the reaction spontaneous, tdeltaS term should be more than deltaH. It should be possible at higher temperatures only.

(18)

For reaction at equilibrium, change in enthalpy = 0, deltaG = 0

(d)

(19)

DeltaG0 = 2 deltaG0(H2O) - 2 deltaG0(H2) - deltaG0(O2)

= 2 (-228.6) - 2 (0) - (0)

= - 457.2 kJ

We know, deltaG0 = - R T lnK

- 457.2 = - 0.008314 * 298 * lnK

lnK = 184.5

K = 1.34 * 1080

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