Given the values of ΔrH∘, ΔrS∘, and T below, determine ΔSuniv.
Part A:
ΔrH∘= 130 kJmol−1 , ΔrS∘=− 263 JK−1mol−1 , T= 291 K .
Part B
ΔrH∘=− 130 kJmol−1 , ΔrS∘= 263 JK−1mol−1 , T= 291 K .
Part C
ΔrH∘=− 130 kJmol−1 , ΔrS∘=− 263 JK−1mol−1 , T= 291 K .
Part D
ΔrH∘=− 130 kJmol−1 , ΔrS∘=− 263 JK−1mol−1 , T= 561 K .

Part A :




This is for a non spontaneous
reaction.
Part B :





This
is for a spontaneous reaction.
Part C :





This
is for a spontaneous reaction.
Part D :





This
is for a non spontaneous reaction.
Given the values of ΔrH∘, ΔrS∘, and T below, determine ΔSuniv. Part A: ΔrH∘= 130 kJmol−1...
Calculate the change in Gibbs energy for each of the sets of ΔrH∘, ΔrS∘, and T. Part A: ΔrH∘=− 90. kJmol−1 , ΔrS=− 150 JK−1mol−1 , T= 302 K Part B: ΔrH∘=− 90. kJmol−1 , ΔrS=− 150 JK−1mol−1 , T= 770 K Part C ΔrH∘= 90. kJmol−1 , ΔrS=− 150 JK−1mol−1 , T= 302 K
A reaction has ΔrH∘=ΔrH∘= -116 kJmol−1 and ΔrS∘=ΔrS∘= 303 JK−1mol−1 . Part A At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?
2Ca(s)+O2(g)→2CaO(s)
ΔrH∘= -1269.8 kJmol−1; ΔrS∘= -364.6 JK−1mol−1
Calculate the Gibbs energy change for the reaction at 24 ∘C.
Express your answer using four significant figures.
2Ca(s) + O2 (g) → 2CaO(s) Δ,Ho =-1269.8 kJ mol 1: Δ,S":-3646 J K 1 mol 1 Part A Calculate the Gibbs energy change for the reaction at 24 °C. Express your answer using four significant figures. kJ mol 1 Submit Previous Answers Request Answer
For a given reaction, ΔrH = +35.5 kJ mol-1 and ΔrS = +83.6 J K-1 mol-1. The reaction is spontaneous ________. Assume that ΔrH and ΔrS do not vary with temperature. A. at all temperatures B. at T < 298 K C. at T < 425 K D. at T > 425 K E. at T > 298 K
Given the values of ΔH∘rxn,ΔS∘rxn, and T below, determine ΔSuniv A) ΔH∘rxn = 125 kJ , ΔS∘rxn = −257 J/K , T= 296 K . B) ΔH∘rxn= −125 kJ, ΔS∘rxn= 257 J/K, T=296 K. C) ΔH∘rxn= −125 kJ, ΔS∘rxn= -257 J/K, T=296 K. D) ΔH∘rxn= −125 kJ, ΔS∘rxn= -257 J/K, T=561 K.
For the decomposition of barium carbonate, consider the following thermodynamic data: ΔrH∘ΔrH∘ 271.5kJ mol−1 ΔrS∘ΔrS∘ 173.8J K−1 mol−1 A: Calculate the temperature in kelvins above which this reaction is spontaneous. Answer:1562K B: Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C: BaCO3(s)→BaO(s)+CO2(g) Answer: 2.73*10^-39 C: When adjusted for any changes in ΔrH and ΔrS with temperature, ΔrG∘(600K)=167kJ mol−1. Calculate the equilibrium constant at 600 K .
Given the values of ΔHorxn, ΔSorxn, and T below, determine ΔSuniv. ΔHorxn= -95 kJ , ΔSorxn= -157 J/K , T= 855 K (HINT: Use the equation on page 876, ΔSuniv = ΔS - ΔH / T ). Group of answer choices -45.9 J/K -111.1 J/K -157 J/K +238.7 J/K -268.11 J/K
Given the values of ΔH∘rxn, ΔS∘rxn, and T below, determine ΔSuniv and whether they are spontaneous or nonspontaneous: A. ΔH∘rxn= 80 kJ , ΔSrxn= 143 J/K , T= 290 K B. ΔH∘rxn= 80 kJ , ΔSrxn= 143 J/K , T= 764 K C. ΔH∘rxn= 80 kJ , ΔSrxn=− 143 J/K , T= 290 K D. ΔH∘rxn=− 80 kJ , ΔSrxn= 143 J/K , T= 396 K
Given the values of ΔH∘rxn , ΔS∘rxn , and T below, determine ΔSuniv . 1. ΔH∘rxn= 83 kJ , ΔSrxn= 152 J/K , T= 308K (spontaneous or nonspontaneous ) 2. ΔH∘rxn= 83 kJ , ΔSrxn= 152 J/K , T= 752 K (spontaneous or nonspontaneous ) 3. ΔH∘rxn= 83 kJ , ΔSrxn=− 152 J/K , T= 308 K (spontaneous or nonspontaneous ) 4. ΔH∘rxn=− 83 kJ , ΔSrxn= 152 J/K , T= 397 K (spontaneous or nonspontaneous )
Use data given below to calculate ΔrS∘ΔrS∘ for each of the reactions. Standard Thermodynamic Quantities for Selected Substances at 25 ∘C∘C Substance ΔrS∘ΔrS∘, JK−1mol−1JK−1mol−1 Cr(s)Cr(s) 23.8 Cr2O3Cr2O3(s)(s) 81.2 CO(g)CO(g) 197.7 CO2(g)CO2(g) 213.8 H2(g)H2(g) 130.7 H2O(g)H2O(g) 188.8 H2O(l)H2O(l) 70.0 HNO3(aq)HNO3(aq) 146.0 N2(g)N2(g) 191.6 N2O4(g)N2O4(g) 304.4 NO(g)NO(g) 210.8 NO2(g)NO2(g) 240.1 O2(g)O2(g) 205.2 SO2(g)SO2(g) 248.2 SO3(g)SO3(g) 256.8 A) 4NO2(g)+2H2O(l)+O2(g)→4HNO3(aq) B) Cr2O3Cr2O3(s)(s) +3CO(g)→2Cr(s)+3CO2(g) C) SO2(g)SO2(g) +12O2(g)→SO3(g) D) N2O4(g)+4H2(g)→N2(g)+4H2O(g)