Acetylene, C2H2, can be converted to ethane, C2H6, by a process known as hydrogenation. The reaction is C2H2(g)+2H2(g)?C2H6(g)

Given the following data, what is the value of Kp for this reaction?

Substance ?G?f
(kJ/mol)
C2H2(g) 209.2
H2(g) 0
C2H6(g) ?32.89

In Part A, we saw that ?G?=?242.1 kJ for the hydrogenation of acetylene under standard conditions (all pressures equal to 1 atm and the common reference temperature 298 K ). In Part B, you will determine the ?G for the reaction under a given set of nonstandard conditions

At 25?C the reaction from Part A has a composition as shown in the table below.

Substance Pressure
(atm)
C2H2(g) 3.85
H2(g) 4.45
C2H6(g)
1.25

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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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

the answer is d.) potassium

d

explanation: