The standard enthalpy of vaporization of Freon-10 is 29.82 kJ/mol at its normal

Question

The standard enthalpy of vaporization of Freon-10 is 29.82 kJ/mol at its normal boiling point ol 77 degree C. What is the change of entropy for 1 mol of liquid Freon-10 when it vaporizes at its norma boiling point? 8.52 times 10-2 J/K 1.04 times 104 J/K 85.2 J/K -85.2 J/K 3.87 times 102 J/K For which of the following processes would delta S degree be expected to be most positive? H2O (l) rightarrow H2O (S) NH3(g) + HCl(g) rightarrow NH4Cl(s) )2te) + 2H2(g)rightarrow 2H2O(g) N2O4(g) rightarrow 2NO2(g) 2NH4NO3(s) rightarrow 2N2(g) + O2(g) + 4H2O(g) For the process H2O (s) rightarrow H2O (l), delta H is + and delta S is + for the reaction. delta H is + and delta S = 0 for the reaction. delta H is - and delta S is - for the reaction. delta H is - and delta S is + for the reaction. delta H is + and delta S is - for the reaction. What is the thermodynamic quantity that provides the criterion for the spontaneity of a chemical reaction? T delta S delta SU delta S delta G delta H What is AG degree at 298 K for the following reaction? I2(g) + Br2(g) rightarrow 2IBr(g); delta H degree = -11.6 kJ; delta Sdegree = 12 J/K at 298 K -8.02 kJ 15.2 kJ -15.2 kJ 3.59 times 103 kJ -3.59 times 103 kJ

Explanation / Answer

1.

delta G= deltaH -T delta S
0= deltaH -T delta S
change in entropy= change in enthalphy/T
          =29820/(273+77)
           =85.2 j/k
ans: C

2.
Ans:E
Since number of moles of gases in product- number of moles of gases in reactant =7 which is largest among all
3.
Ans:A
It requires heat so delta H is positive.
Entropy ofliquid is more then entropy of solid, so delta S is positive

4.
Ans:D
It is a known fact that delta G= negative implies reaction is spontantaneous

5.
delta G= deltaH -T delta S
   =-11.6 -298*0.012
= -15.2 kJ
ans:C

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