Question

Using values from Appendix C of your textbook, calculate the value of K_eq at 298 K for each of the following reactions:

(a) 4 NH₃(g) + O₂(g) 2 N₂H₄(g) + 2 H₂O(l)

K_eq =  .



(b) CH₄(g) + 4 Cl₂(g) CCl₄(l) + 4 HCl(g)

K_eq =  .



(c) 6 C(graphite) + 6 H₂O(l) C₆H₁₂O₆(s)

K_eq =  .

Answer 1

Answer -

Given,

Temperature = 298K

K_eq = ?

Useful information,

Species	$\Delta$G$\degree$f(kJ/mol)
NH3(g)	-16.45
O2(g)	0
N2H4(g)	+120.3
H2O(l)	-237.129
CH4(g)	-50.72
Cl2(g)	0
CCl4(l)	-65.21
HCl(g)	-95.299
C(graphite)(s)	0
C6H12O6(s)	-304.3

a)

4 NH₃(g) + O₂(g) 2 N₂H₄(g) + 2 H₂O(l)

We know that,

$\Delta$G^$\degree$ = -RTlnK_eq

K_eq = e^{-$\Delta$G$\degree$/RT} ---------------------A

where,$\Delta$G^$\degree$ = Gibbs Free Energy

R = 8.314 Jmol^-1K^-1

T = Temperature

Also,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 2 * $\Delta$ G^$\degree$_f(N₂H₄(g) + 2 * $\Delta$ G^$\degree$_f(H₂O(l)) - 4 *$\Delta$G^$\degree$_f(NH₃(g)) - 1* $\Delta$ G^$\degree$_f(O₂(g))

$\Delta$G^$\degree$_rxn = [2 * (+120.3 kJ/mol)] + [2 * (-237.129 kJ/mol)] - [4 *(-16.45 kJ/mol)] - [1* (0 kJ/mol)]

$\Delta$G^$\degree$_rxn = -167.858 kJ/mol or -167858 J/mol [1kJ/mol = 1000J/mol]

Put this in A,

K_eq = e^{-(-167858 J/mol/(8.314 Jmol-1K-1 * 298K)}

K_eq = e^67.75

K_eq = 2.65 * 10²⁹ [Answer]

(b)

CH₄(g) + 4 Cl₂(g) CCl₄(l) + 4 HCl(g)

We know that,

$\Delta$G^$\degree$ = -RTlnK_eq

K_eq = e^{-$\Delta$G$\degree$/RT} ---------------------A

where,$\Delta$G^$\degree$ = Gibbs Free Energy

R = 8.314 Jmol^-1K^-1

T = Temperature

Also,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 1 * $\Delta$ G^$\degree$_f(CCl₄(l) + 4 * $\Delta$ G^$\degree$_f(HCl(g)) - 4 *$\Delta$G^$\degree$_f(Cl₂(g)) - 1* $\Delta$ G^$\degree$_f(CH₄(g))

$\Delta$G^$\degree$_rxn = [1 * (-65.21 kJ/mol)] + [4 * (-95.299 kJ/mol)] - [4 *(0 kJ/mol)] - [1* (-50.72 kJ/mol)]

$\Delta$G^$\degree$_rxn = -395.686 kJ/mol or -395686 J/mol [1 kJ/mol = 1000J/mol]

Put this in A,

K_eq = e^{-(-395686J/mol/(8.314 Jmol-1K-1 * 298K)}

K_eq = e^159.71

K_eq = 2.29 * 10⁶⁹ [Answer]

(c)

6 C(graphite) + 6 H₂O(l) C₆H₁₂O₆(s)

We know that,

$\Delta$G^$\degree$ = -RTlnK_eq

K_eq = e^{-$\Delta$G$\degree$/RT} ---------------------A

where,$\Delta$G^$\degree$ = Gibbs Free Energy

R = 8.314 Jmol^-1K^-1

T = Temperature

Also,

$\Delta$G^$\degree$_rxn = $\sum$ n * $\Delta$ G^$\degree$_f(products) - $\sum$ n * $\Delta$ G^$\degree$_f(reactants)

where, n = stiochiometric coefficient

$\Delta$G^$\degree$_rxn = 1 * $\Delta$ G^$\degree$_f(C₆H₁₂O₆(s)) - 6 *$\Delta$G^$\degree$_f(H₂O(l)) - 6* $\Delta$ G^$\degree$_f(C(graphite))

$\Delta$G^$\degree$_rxn = [1 * (-304.3 kJ/mol)] - [6 * (-237.129 kJ/mol)] - [6 *(0 kJ/mol)]

$\Delta$G^$\degree$_rxn = 1118.474 kJ/mol or 1118474 J/mol [1 kJ/mol = 1000J/mol]

Put this in A,

K_eq = e^{-(1118474J/mol/(8.314 Jmol-1K-1 * 298K)}

K_eq = e^-451.43

K_eq = 8.84 * 10^-197 [Answer]