For an equilibrium reaction: A + B C + D the following graph was obtained. In K

Question

For an equilibrium reaction: A + B C + D the following graph was obtained. In K = - Delta H degree/RT + Delta S degree/R Given the above relationship, use the graph to identify which of the following statements are "true" or "false" in this specific case. All statements refer to the forward reaction (proceeding left to right, as given) corresponding to the equilibrium constant k. The reaction is spontaneous at all temperatures. As the reaction proceeds, the system becomes more disordered. G degree is positive at all temperatures. The reaction is endothermic. As the temperature is decreased, the equilibrium position shifts in favour of the products.

Explanation / Answer

from the plot of lnK vs 1/T, the slope gives -deltaH/R and intecept gives deltaS/R

from the plot given=(-95-20)/0.01= -deltaH/R, deltaH= 8.314*11500 J/mole= 95611 j//mole

deltaS/R= intercept= 20 J/mole.K

deltaS= 20*8.314 J/mole.K= 166.3 J/mole.K

deltaG= deltaH-T*deltaS= 95611-166.3T

for a reaction to be spontaneous at all temperatures (K), deltaG has to be -ve.

but for deltaG to be -ve,   arrive at value at which deltaG=0

95611-166.3T=0

T= 95611/166.3 =575K, avove 575K only the reaction is spontaneous. ( False)

since entropy change is +ve and postive entropy change indicates increased disorder, as the reaction proceeds, there will be more disorder. ( True)

deltaG0= deltaHo-T0*deltaSo = 95611-166.3*298 =46054 j/mole.K so +ve at only 298K (0 refer to standard conditions of pressure and temperature)

deltaH is +ve indicating the reaction is endothermic ( True).

Since the reaction is enothermic, as temperature is decreased, as per Lechatlier principle, the reaction proceeds in a direction where there is an increase in temperature as so to compensate the effect of decrease in temperature.

Hence the reaction proceeds backwards. (False)

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