Question

Which of the following processes have a AS> 0? CH3OH(l) → CH3OH(s) ON 2(g) + 3 H2(g) → 2 NH3(g) CH4(g) + H2O(g) → CO(g) + 3 H2(g) ON a2CO3(s) + H2O(g) + CO2(g) → 2 NaHCO3(s) All of the above processes have a DS>0

2. For the following example, identify the following.

H₂O(l) → H₂O(s)

question 2 options

	a negative ΔH and a negative ΔS
	a positive ΔH and a negative ΔS
	a negative ΔH and a positive ΔS
	a positive ΔH and a positive ΔS
	It is not possible to determine without more information. 3. Calculate ΔS°rxn for the following reaction. The S° for each species is shown below the reaction. C2H2(g) + H2(g) → C2H4(g) S°(J/mol∙K) 200.9 130.7 219.3 Question 4 options: +112.3 J/K +550.9 J/K -112.3 J/K +337.1 J/K -550.9 J/K 4. Calculate ΔS°rxn for the following reaction. The S° for each species is shown below the reaction. C2H2(g) + 2 H2(g) → C2H6(g) S°(J/mol∙K) 200.9 130.7 229.2 Question 5 options: +303.3 J/K +560.8 J/K -102.4 J/K -233.1 J/K 229.2 J/K 5. Calculate the ΔG°rxn using the following information. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1 Question 6 options: -162.5 kJ +51.0 kJ -54.5 kJ +171.1 kJ -87.6 kJ PLEASE SHOW WORK FOR ME!!!

Answer 1

Entropy is degree of disorder. When ever, the solid becomes a liquid, the degree of disorder increases and when ever a liquid becomes a gas also entropy increases. In case of gaseous reaction, entropy increases or decreases depending upon the no of moles of reactants and products. If moles of products are more than moles of reactants, entropy change increases while if the moles of products are less than moles of reactants, entropy change decreases.

For the reaction, CH3OH(l) --->CH3OH(s), entropy change decreases since q liquid is becoming a solid

For the reaction, N2(g)+3H2(g) ---->2NH3(g), the moles of products are less than moles of reactants. So entropy change decreases.

For the reaction, CH4(g)+ H2O(g)---->Co(g)+3H2, moles pf products are more than moles of reactants. So entropy change increases.

For the reaction Na2CO3(s)+H2O(g)+CO2(g)---->2NaHCO3(s)

Moles of gaseous products are more on reactants side. So entropy change decreases.

2. H2O(l)------>H2O(s),a liquuid is converted to solid by removing heat of fusion. Since heat is removed from system, enthalpy change is -ve. Since a liquid becomes solid, entropy change decreases. hence negative deltaH and negative deltaS.

3. For the reaction, C2H2(g)+ H2(g) ------>C2H4(g), entropy change= sum of entropy of products- sum of entropy of reactants

entropy change= 1* entropy of C2H4- (1* entropy of C2H2+1* entropy of H2), 1,1, 1 are coefficients o C2H4, C2H2 and H2 respectively.

=219.3- (200.9+130.7)=-112.3 J/K ( C is correct)

4. Calculate ΔS°rxn for the following reaction. The S° for each species is shown below the reaction.

for the reaction, entropy change= 229.2-( 200.9+2*130.7) =-233.1 J/K ( D is correct)

5.

Calculate the ΔG°rxn using the following information.

2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l)

deltaG= 1* deltaG of H2O+3* deltaG of NO2- (2* deltaG of HNO3+1* deltaG of NO)

=-237.1+3*(51.3)- {2*-110.9+1*87.6)
ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1=51 KJ ( B is correct)