Respuesta :
Answer:
a) Kp 0.151 at 1000K, Kp at 298.15 is 2.15x10⁻¹⁸
b) ΔG = 100.8 kJ/mol
Explanation:
For the reaction:
C2H6(g)⇌C2H4(g)+H2(g)
Kp is defined as:
[tex]Kp = \frac{P_{H_{2}}P_{C_2H_4}}{P_C_2H_6}[/tex]
a) It is possible to obtain pressure of each compound using mole percent of gases, that is:
[tex]Kp = \frac{0.266*0.266}{0.468}[/tex] = 0.151 at 1000K
Here, you can use:
[tex]ln\frac{K2}{K1}=-\frac{dH}{R} (\frac{1}{T2} -\frac{1}{T1} )[/tex]
Where K is K of reaction, dH is ΔH, R is gas constant (8.314 J/molK) and T are temperatures.
Replacing:
[tex]ln\frac{K2}{0.151}=-\frac{137000J/mol}{8.314J/molK} (\frac{1}{298.15} -\frac{1}{1000K} )[/tex]
ln K2 /15.1 = 38.8
K2 / 0.151 = 1.42x10⁻¹⁷
K2 = 2.15x10⁻¹⁸
b) ΔG of reaction at 298.15K is:
ΔG = -RT ln K
Where R is gas constant (8.314J/molK), T is temperature (298.15K), K is Kp (2.15x10⁻¹⁸)
ΔG = -8.314J/molK×298.15K ln 2.15x10⁻¹⁸
ΔG = 100.8 kJ/mol
Answer:
1) Kp at 1000 K = 0.151
2) Kp at 298.15 K = 2.15*10^-18
3) ΔG = 1.0 *10^5 J/mol
Explanation:
Step 1: Data given
Temperature = 1000 K
The pressure stays cosntant at 1.00 bar
the composition of the mixture in mole percent is:
H2(g): 26.6 %
C2H4(g): 26.6 %
C2H6(g): 46.8 %.
Step 2: The balanced equation
C2H6(g)⇌C2H4(g)+H2(g)
Step 3: Calculate Kp at 1000 K.
Kx = ((pC2H4)*pH2)/ (pC2H6)
Kx = (0.266 * 0.266) / (0.468)
Kx = 0.151
Step 4: If ΔH of reaction = 137 kJ/mol, calculate the value of Kp at 298.15 K.
Kp(1000) = Kx * P/P° since the pressure stays constant at 1 bar
Kp (1000K) = 0.151 * 1
Kp(1000K) = 0.151
ln Kp(298.15K) = ln Kp (1000K) - ΔH / R((1/298.15) - (1/1000))
ln Kp(298.15K) = -1.89 (137 *10^3 J/mol / 8.314 J/mol*K) ((1/298.15) - (1/1000))
ln Kp(298.15K) = -40.68
Kp(298.15K) = 2.15*10^-18
Step 5: Calculate ΔG of reaction for this reaction at 298.15 K.
ΔG(298.15K) = -R*T*ln Kp(298.15K)
ΔG(298.15K) = -8.314 J/mol*K * 298.15 * -40.68
ΔG(298.15K) = 100838 K/mol = 1.0 * 10^5 J/mol
