In an experiment, 1.000 atm of H2(g) in a 5.00 L container at 25.00° standard st

Question

In an experiment, 1.000 atm of H2(g) in a 5.00 L container at 25.00° standard state conditions with a stoichiometric quantity of 02(g) to form water vapor. Determine the entropy change. (answer in J/K) Substance S(J/mol·K) H2(g) 02(g) H20(g) H2O() C was reacted under 130.58 205.0 188.83 69.95 Determine the standard entropy change of the universe when 0.4 mol of NLNO, is produced H'(aq) + NO3(aq) NH4NO3(s), given the following information. Is the reaction spontaneous under standard conditions? (answer in J/K) Substance Hf/kJ/mol) S(J/mol. K) NH4NO3(s) -365.6 NH (aq) -132.5 NO, (aq)-205.0 151.1 113.4 146.4

Explanation / Answer

The reaction is H2(g)+ 0.5O2(g) ------àH2O(g)

Entropy change= entropy change of products- entropy change of reactants

Moles of H2 in the container = PV/RT, P= pressure =1atm, V= 5L, R =0.0821 L.atm/mole.K

T= 25deg.c= 25+273= 298K, n= 1*5/(0.0821*298)= 0.20 moles

Moles of O2= 0.2*0.5=0.1, moles of H2O= 0.2

Hence entropy change = 0.2* 188.83- (130.58*0.2+0.1*205) J/K=-8.85 J/K

2.

Moles od NH4+ used =mole of NO3- used =0.4

Entropy change= entropy change of products- entropy change of reactants

Entropy change= 0.4* 151.1-(0.4*113.4+0.4*146.4) =-43.48 J/K

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.