could someone show how to get these answers? 1st question answer is +198.8 J/K 1
ID: 857088 • Letter: C
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could someone show how to get these answers?
1st question answer is +198.8 J/K
1st question answer is +198.8 J/KA.mol 2nd question answer is 89.0 J/KA.mol 3rd and 4th question answer provided, just need elaboration on why. Find. S degree at 25 degree C for the reduction of PbO(s), 2PbO(s) + C(s) . 2 Pb(s) + CO2(g) S degree (J/K.mol) PbO(s) 69.45 C(s) 5.7 Pb(s) 64.89 CO2(g) 213.6 HI has a normal boiling point of -35.4 degree C, and its . Hvapor is 21.16 kJ/mol. Calculate the molar entropy of vaporization (. Svap). Why is this consistent with a reaction at equilibrium? . G = 0, Q = K Why is this consistent with a reaction that proceeds spontaneously in the reverse direction (assume all variables are in terms of the forward direction only)? . G > 0, Q > KExplanation / Answer
1] delta S of reaction = delta S* of products - delta S* of reactants [including coefficients]
delta S of reaction = [2* delta S of Pb + delta S of CO2] - [2*delta S of PbO + delta S of C]
keeping all values ;
delta S of reaction = 198.78J/K
2] at vapourization it is in equilibrium
delta G = delta H - T delta S
delta G = 0 in equilibrium
therefore delta S = delta H /T
keeping all values T in Kelvin
delta S = 89 J/K
3] A simple relationship between Kc and the reaction quotient, known as Qc, can help. The reaction quotient, Q, expresses the relative ratio of products to reactants at a given instant. Using either the initial concentrations or initial activities of all the components of the reaction, the progression of an reaction can easily be determined.
delta G = delta G* - RTlnK
since delta G = 0
if delta G = 0 it means it is in equilibrium;
then Q= K
4] When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants.
if reverse reaction favours it means delta G > 0 for the reaction [since forward reaction is non spontaneous and reverse reaction favors]
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