Respuesta :
Answer:
part A = K = 2.7*10^42
Part B = ΔG = -252284 J = -252.3 kJ
Explanation:
Step 1: Data given
Step 2: The balanced equation
C2H2(g)+2H2(g) ⇄ C2H6(g)
Δ Gf = (1* GfC2H6) - (1*GfC3H2)
Δ Gf = -209.2 -32.89
Δ Gf = - 242.09 kJ/mol = -242090 J/mol
Step 3 Calculate K
Δ Gf = -RT ln K
⇒Δ Gf = -242090 J/mol
⇒R = 8.314 J/mol*K
⇒T = the temperature = 298 K
⇒ K = the equilibrium constant
-242090 = - (8.314)(298) lnK
K = 2.7*10^42
PArt B:
Δ Gf = -242.1 kJ= -242100 J
Q = pressure products / pressure reactants
Q = (pC2H6) / (pH2² * pC2H2)
Q = 1.25 / (4.45²*3.85)
Q= 0.0164
ΔG = ΔG° + RT ln Q
ΔG = -242100 + 8.314 * 298 * ln 0.0164
ΔG = -252284 J = -252.3 kJ
Part A : The value of Kp for the reaction is : 2.7 * 10⁴²
Part B : The value of ΔG = -252.3 kJ
Chemical reaction : C₂H₂(g) + 2H₂(g) ⇄ C₂H₆(g)
ΔGf = -209.2 -32.89 = -242090 J/mol
R = 8.314 J/mol*K
T = 25° C + 273 = 298 K
A) Determine the value of Kp
Apply the equation below
Δ Gf = -RT ln k ---- ( 1 )
∴ ln k = ( ΔGf ) / ( - RT )
= ( - 242090 ) / ( - 8.314 * 298 )
= 97.713
∴ K = 2.7 * 10⁴²
B) Determine the value of ΔG
ΔG = ΔG° + RT ln Q ----- ( 2 )
Where : ΔG° = -242100 J, R = 8.314 , Q = 0.0164
Q = 1.25 / (4.45² * 3.85) = 0.0164.
Insert values into equation ( 2 ) above
ΔG = -242100 + ( 8.314 * 298 ) ln 0.0164
= -252284 J
= -252.3 kJ
Hence we can conclude that The value of Kp for the reaction is : 2.7 * 10⁴² and The value of ΔG = -252.3 kJ.
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