Answer:
Kc = 1.4752
Explanation:
⇒ Kc = ( [ CO2(g) ] * [ H2 ]∧4 ) / ( [ CH4 ] * [ H2O ]² )
∴ [ CO2 ] = ( 8.01 g * ( mol / 44.01 g ) ) / 1.5 L
⇒ [ CO2 ] = 0.1213 M
⇒ [ H2O ] = 0.877 mol / 1.5 L = 0.585 M
⇒ [ CH4 ] = 0.380 mol / 1.5 L = 0.253 M
⇒ [ H2 ] = (( 0.380 mol CH4 ) * ( 4 mol H2 / mol CH4 )) / 1.5 L = 1.013 M
⇒ Kc = (( 0.1213 ) * ( 1.013 )∧4 ) / (( 0.253 ) * ( 0.585 )² )
⇒ Kc = 1.4752
The value of Kc is 1.4752
Kc is the ratio of equilibrium product concentrations to equilibrium reactant concentrations.
[tex]\rm CH_4 (g) + 2H_2O (g) < - > CO_2 (g) + 4H_2 (g)[/tex]
[tex]Kc = \dfrac{( [ CO_2(g) ] \times [ H_2 ]^4 )}{ ( [ CH_4 ] \times [ H_2O ]^2 )}[/tex]
For CO2
[tex][ CO_2 ] =\dfrac{( 8.01 g \times ( mol / 44.01 g ) ) }{1.5 }[/tex]
[tex]\rm [ CO_2 ] = 0.1213 M[/tex]
For H2O
[tex]\rm [ H_2O ] = \dfrac{0.877 mol }{ 1.5 L} = 0.585 M[/tex]
Fore CH4
[tex][ CH_4 ] = \dfrac{0.380 mol}{ 1.5 L } = 0.253 M[/tex]
For H₂
[tex]\rm [ H_2 ] = \dfrac{(( 0.380 mol CH4 ) \times ( 4 mol H_2 / mol CH_4 ))}{1.5 L} = 1.013 M\\\\ Kc = \dfrac{(( 0.1213 ) \times ( 1.013 )^4 ) }{(( 0.253 )\times ( 0.585 )^2 )} \\\\ Kc = 1.4752[/tex]
Thus, the value of Kc is 1.4752
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