Question

Reactive Metal Fires In the chemistry lab, one often has to work with very reactive metels such as sodium end lithium. Unfortunately, these same metals can react vigorously with water or oxygen and cause res. Given the following information: Alyamon AH (J/mol LI2O(5) -597.9 Lila) -278.5 Na 2015) -415 Nal(aq) -240 K2015) -361 K*(0) -251 Cog) -110.5 CO2(9) -393.5 HO/V) -296 OH(aq) -230 CC140 -135 SiOs) -911 LICIOS) -409 Nack(s) -411 KCI(S) -425 chao) -167 Calculate AH for the following reactions: 4 Na(s) 1 0,0) + z Na 20/5) 2 Submit AnswerTiles 0/5 Na(s) ICO(0) Na2O(5) cola! Submit Answer Tries 0/5 2Na(s) + 2H2010 2NaOH(aq) + H»(s) its

Answer 1

Consider a reaction, 4 Na (s) + O ₂ (g) $\rightarrow$ 2 Na₂O (s)

The standard enthalpy change of reaction is given as

r H ⁰ = H ⁰_f ( products ) -   H ⁰_f ( reactants )

H ⁰_f (products) = 2 H ⁰ _f Na₂O (s) = 2 ( - 416 kJ /mol ) = - 832 kJ / mol

H ⁰_f ( reactants ) = 4 ( H ⁰ _f Na (s) ) + H ⁰ _f O ₂ (g)

H ⁰_f ( reactants ) = 4 ( 0 kJ /mol ) + 0 kJ / mol = 0 kJ / mol

$\therefore$ r H ⁰ = ( - 832 kJ /mol ) - o kJ / mol = - 832 kJ / mol

ANSWER : r H ⁰ = - 832 kJ / mol

Consider a reaction, 2 Na (s) + CO ₂ (g) $\rightarrow$ Na₂O (s) + CO (g)

The standard enthalpy change of reaction is given as

r H ⁰ = H ⁰_f ( products ) -   H ⁰_f ( reactants )

H ⁰_f (products) = H ⁰ _f Na₂O (s) + H ⁰ _f CO (g)

= ( - 416 kJ /mol ) + ( -110.5 kJ /mol )

= - 526.5 kJ / mol

H ⁰_f ( reactants ) = 2 H ⁰ _f Na (s) + H ⁰ _f CO ₂ (g)

= 2 ( 0 kJ / mol ) + ( - 393.5 kJ /mol)

= - 393.5 kJ / mol

$\therefore$ r H ⁰ = - 526.5 kJ / mol - (- 393.5 kJ / mol)

= - 526.5 kJ / mol + 393.5 kJ / mol

= - 133 kJ / mol

ANSWER : r H ⁰ = - 133 kJ / mol

Consider a reaction, 2 Na (s) + 2 H₂O (l) $\rightarrow$ 2 NaOH (aq) + H ₂ (g)

The standard enthalpy change of reaction is given as

r H ⁰ = H ⁰_f ( products ) -   H ⁰_f ( reactants )

H ⁰_f (products) =2 H ⁰ _f NaOH (aq) + H ⁰_f H ₂ (g)

= 2 ( - 470.11 kJ / mol ) + 0 kJ/ mol ( value taken from literature)

= - 940.22 kJ / mol

H ⁰_f ( reactants ) = 2 H ⁰ _f Na( s) + 2 H ⁰_f H₂O (l)

= 2 ( 0 kJ / mol ) + 2 ( - 286 kJ / mol )

= - 572 kJ / mol

$\therefore$   r H ⁰ = - 940.22 kJ / mol - ( - 572 kJ / mol)

= - 940.22 kJ / mol + 572 kJ / mol

= - 368.22 kJ /mol

ANSWER : r H ⁰ = -368.22 kJ / mol