Consider the following reaction at 298 K: C(graphite)+2CL(g)------> CCl4(l) delta H=-139 kj Calculate delta S sys...

Consider the following reaction at 298 K: C(graphite)+2CL(g)------> CCl4(l) delta H=-139 kj

Calculate

delta S sys

delta S surr

delta S univ

Expert Solution

entropy data ( J/mole.K) : C ( graphite) : 5.7 Cl2 (g):223.81: CCl4 : 216.4

the reaction is C(graphite)+ 2Cl2(g) ------>CCl4(l)

entropy change = Sum of entropy of products- { sum of entropy of reactants}

1, 2 and 1 are coefficients of CCl4(l), Cl2(g), C(graphite)

= 1*216.4-{1* 5.7+2*223.81}=-236.92 J/mole.K

Since during the reaction, enthalpy change is -ve suggesting that heat is evolved and surroundings gain entropy

entropy change of surroundings= +139*1000/298.15=466.21 J/Mole.K

entropy change of universe= entropy change of surroundings+ entropy change because of reaction =466.21-236.92 =229.29 J/mole.K

queen_honey_blossom answered 2 years ago

Consider the following reaction at 298 K: 4Al(s) + 3O2(g) ==> 2Al2O3(s) Delta H= -3351.4 kJ/mol...

Consider the following reaction at 298 K: 4Al(s) + 3O2(g) ==> 2Al2O3(s) Delta H= -3351.4 kJ/mol Calculate: a. Delta Ssystem = _______J/mol*K b. Delta Ssurroundings = _______J/mol*K c. Delta S universe = _________J/mol*K

Use delta H f and S to calculate delta G rxn (delta g not sys) at...

Use delta H f and S to calculate delta G rxn (delta g not sys) at 25 C for the reaction below: 4KClO3 (s) --> 4KClO4 (s) + KCL (s) Delta H not f KCLO3 = -397.7 kj/mol; KCLO4 = -432.8 kJ/mol; KCL = -436.7 kJ/mol S KCLO3 = 143.1 kJ/mol ; KCLO4= 151.0 kJ/mol; KCL = 82.6 kJ/mol

Calculate the standard cell potential Delta G of the following reaction: Cd(s)+Cl2(g) --> 2Cl- + Cd2+(s)

A reaction has and △H°298 = 151 kJ/mol and △S°298 = 286 J /mol K at...

A reaction has and △H°298 = 151 kJ/mol and △S°298 = 286 J /mol K at 298 K. Calculate △G in kJ/mol.

Consider the reaction: 2 NO2(g) → N2O4(g) Calculate ΔG (in kJ/mol) at 298°K if the equilibrium...

Consider the reaction: 2 NO2(g) → N2O4(g) Calculate ΔG (in kJ/mol) at 298°K if the equilibrium partial pressures of NO2 and N2O4 are 1.337 atm and 0.657 atm, respectively.

Part 1. Consider the following reaction at 298 K: 2H2(g)+O2(g)---->2H2O(g) ΔH=-483.6 kj/mol Calculate the following quantities:...

Part 1. Consider the following reaction at 298 K: 2H2(g)+O2(g)---->2H2O(g) ΔH=-483.6 kj/mol Calculate the following quantities: ΔSsys=____J(molxK) ΔSsurr=____J(molxK) ΔSuniv=____J (molxK) Part 2. For a particular reaction, ΔH = 168.1 kJ/mol and ΔS = -55.8 J/(mol·K). Calculate ΔG for this reaction at 298 K. Is this system spontaneaus as written, Is it in the reverse direction, Or is it at equilibrium?

Calculate ?H (298 K) per gram of fuel units of kJ*g^-1 (exclude oxygen) a.) H2(g) +...

Calculate ?H (298 K) per gram of fuel units of kJ*g^-1 (exclude oxygen) a.) H2(g) + O2(g) = H2O(g) b.) CH4(g) + 2O2(g) = CO2(g) + 2H2O(g) c.) CH3OH(l) +3/2 O2(g) = CO2(g) + 2H2O(g) d.) C6H14(g) + 9 1/2 O2(g) = 6CO2(g) + 7H2O(g)

Use ΔGº =ΔHº -TΔSº to calculate ΔG (in kJ) at 298 K for : 2CO2(g) +4H2O(l)...

Use ΔGº =ΔHº -TΔSº to calculate ΔG (in kJ) at 298 K for : 2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g) If the above reaction could be done at 2081 K, what would be your estimate for ΔGº (in kJ) at this elevated temperature? Use ΔGº =ΔHº -TΔSº and assume ΔHº and ΔSº are independent of temperature. (The º is included because it is still for standard conditions, that is, 1 atm for gases and 1 molar for concentrations.)

Consider the reaction at 298 K SO2(g) + 2H2S(g) ? 3S(s) + 2H2O (g) The ?G

A.) The standard enthalpy change for the following reaction is 359 kJ at 298 K. PbCl2(s)...

A.) The standard enthalpy change for the following reaction is 359 kJ at 298 K. PbCl2(s) ----> Pb(s) + Cl2(g) ΔH° = 359 kJ What is the standard enthalpy change for this reaction at 298 K? Pb(s) + Cl2(g) ----> PbCl2(s) __________ kJ B.) The standard enthalpy change for the following reaction is -602 kJ at 298 K. Mg(s) + 1/2 O2(g) ----> MgO(s) ΔH° = -602 kJ What is the standard enthalpy change for the reaction at 298 K? 2...

Question