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1001 MCAT Chemistry- Question 433 and 540
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umghai
#1 Posted : Tuesday, July 19, 2011 1:56:55 AM
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433) According to the classroom companion, standard conditions occur at 25 C, so why do they choose a temperature of 0. Is it for simplicity to solve for enthalpy?

540) I do not know how to solve for x as it creates a binomial factoring due to FOIL rules when solving for x.
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selenahuang
#2 Posted : Wednesday, July 20, 2011 12:42:43 AM
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433) This question does not really utilize the temperature at which the reaction is occurring. The graph is plotted for the STANDARD Gibb's free energy change. The 0K is set as the standard Gibb's free energy is defined as (delta)G(0) = (delta)H(0) - T(delta)S(0). If you artificially set the T = 0K, the TS part is 0, so at this point on the graph, G = H(0).

540) Hopefully after constructing an ICE table you have arrived at 1.3 x 10^(-6) = x(0.1+2x)^2. At this point, it's important to realize that x << 0.1. As a result, the expression can be simplified to 1.3 x 10^(-6) = x(0.1)^2. Thus x = 1.3 x 10^(-4)
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