Question

Consider the reaction Ca(OH)2(aq) + 2HCl(aq) +CaCl (s) + 2H2O(1) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.26 moles of Ca(OH)2(aq) react at standard conditions. AS surroundings = J/K

Answer 1

Following is the - complete Answer -&- Explanation: for the given: Question Set...in...typed format...

$\Rightarrow$Answer:

1. question - 1:    $\Delta$ S^o_surrounding =   228.86 J/K
2. question - 2:    $\Delta$ S^o_surrounding =   1519.376 J/K

$\Rightarrow$Explanation:

Following is the complete: Explanation: for the above: Answers...in...typed format...

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$\Rightarrow$Question - 1:

• Given:

1. balanced chemical equation: Ca(OH)₂ (aq) + 2 HCl (aq)  $\rightleftharpoons$ CaCl₂ (s) + 2H₂O (l) -------- Equation - 1
2. Temperature during the reaction: T = 298 K ( Kelvin )
3. Number of moles, of Ca(OH)₂ = 2.26 moles

• ​​​​​​​Step - 1:

​​​​​​​We know, from Equation -1 , the following molar ratio, between products and reactants:

$\Rightarrow$  molar ratio: Ca(OH)₂ : HCl :  CaCl₂ : H₂O = 1 : 2 : 1 : 2  

Therefore: if moles, of Ca(OH)₂ = 2.26 moles, we will have the following:

1. Number of moles of HCl =  ( 2.26 x 2 ) =  4.52 moles
2.   Number of moles of CaCl₂ =  ( 2.26 x 1 ) =  2.26 moles
3.   Number of moles of H₂O =  ( 2.26 x 2 ) =  4.52 moles

• ​​​​​​​Step -2:

​​​​​​​We know the following:

Compounds	Standard enthalpy of formation ( i.e. $\Delta$ Hfo ) [=] kJ / mol
Ca(OH)2 (aq)	-1002.82 kJ / mol
HCl (aq)	-167.2 kJ / mol
CaCl2 (s)	-795.8 kJ / mol
H2O	-285.8 kJ/mol

• Step - 3:

​​​​​​​Therefore: the value of the change in enthalpy: of the reaction: Equation - 1: will be: the following:

$\Rightarrow$ $\Delta$ H_rxn.^o = [ (2.26 moles) x ( $\Delta$ H_f^o[CaCl₂ ] ) + (4.52 moles) x ($\Delta$​​​​​​​H_f^o[H₂O] ) ] - [ (2.26 moles) x (  $\Delta$H_f^o[Ca(OH)₂ ] ) + ( 4.52 moles) x (  $\Delta$H_f^o[HCl ] ) ]

$\Rightarrow$   $\Delta$ H_rxn.^o =   [ (2.26 moles) x ( -795.8 kJ / mol ) + (4.52 moles) x (-285.8 kJ/mol ) ] - [ (2.26 moles) x (  -1002.82 kJ / mol ) + ( 4.52 moles) x ( -167.2 kJ / mol ) ]

$\Rightarrow$$\Delta$H_rxn.^o = - 68.2 kJ = - 68200 J ( Joule )

• Step - 4:

​​​​​​​Therefore: we will get the following:

$\Rightarrow$    $\Delta$ S^o_surrounding = - $\Delta$ H_rxn.^o / T =   - ( - 68200 J ) / ( 298 K ) = 228.86 J/K

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$\Rightarrow$Question - 2:

• Given:

1. ​​​​​​​balanced chemical equation: 2CO(g) + O₂(g) $\rightleftharpoons$ 2 CO₂ (g) -------- Equation - 2
2. Temperature during the reaction: T = 298 K ( Kelvin )
3. Number of moles, of CO(g) = 1.60 moles

• ​​​​​​​Step - 1:

​​​​​​​We know, from Equation -2 , the following molar ratio, between products and reactants:

$\Rightarrow$  molar ratio: CO : O₂ : CO₂ = 2 : 1 : 2

Therefore: if moles, of CO(g) = 1.60 moles, we will have the following:

1. Number of moles of O₂ (g) =  ( 1.60 / 2 ) =  0.8 moles
2.   Number of moles of CO₂ (g) =  ( 1.60 x 1 ) = 1.60 moles

• ​​​​​​​Step -2:

​​​​​​​We know the following:

Compounds	Standard enthalpy of formation ( i.e. $\Delta$ Hfo ) [=] kJ / mol
CO (g)	-110.525 kJ/mol
O2 (g)	0.0 kJ / mol
CO2 (g)	-393.509 kJ/mol

• Step - 3:

​​​​​​​Therefore: the value of the change in enthalpy: of the reaction: Equation - 1: will be: the following:

$\Rightarrow$ $\Delta$ H_rxn.^o = [ (1.60 moles ) x ( $\Delta$ H_f^o[ CO₂ (g) ] ) ] - [ (1.60 moles) x ( $\Delta$ H_f^o[CO(g) ] ) + ( 0.8 moles ) x ($\Delta$H_f^o[O₂ ] ) ]

$\Rightarrow$   $\Delta$ H_rxn.^o =   [ (1.60 moles ) x (-393.509 kJ/mol ) ]  - [ (1.60 moles) x ( -110.525 kJ/mol ) + ( 0.8 moles ) x ( 0.0 kJ / mol ) ]

$\Rightarrow$$\Delta$H_rxn.^o = - 452.77 kJ  = - 452774 J ( Joule )

• Step - 4:

​​​​​​​Therefore: we will get the following:

$\Rightarrow$    $\Delta$ S^o_surrounding = - $\Delta$ H_rxn.^o / T =   - ( - 452774 J ) / ( 298 K ) =  1519.376 J / K

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