Question

assignmentsession Locator assignment take [Review Topical Reference Use the References to es important values if needed for this question. Consider the reaction CO(g) + Cl (2)— COCI(8) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K. ANSWER: Submit Answer Retry Entire Group 2 more group attempts remaining

Answer 1

Answer -

Given,

Temperature = 298.15 K

Keq = ?

CO (g) + Cl₂ (g)  $\rightarrow$ COCl₂ (g)

Some Useful information,

Species	$\Delta$G$\degree$f (kJ/mol)
CO (g)	-137.168
Cl2 (g)	0
COCl2 (g)	-204.6

We know that,

$\Delta$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 COCl₂ (g)) - 1 * $\Delta$ G^$\degree$_f (CO (g)) - 2 *$\Delta$G^$\degree$_f (Cl₂ (g))

Put the values,

$\Delta$G^$\degree$_rxn = 1 * -204.6 kJ/mol) - 1 * -137.168 kJ/mol) - 2 * 0 kJ/mol)

$\Delta$G^$\degree$_rxn = -67.432 kJ /mol

We know that,

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

^{where, Keq = Equilibrium constant}

$\Delta$G^$\degree$ = Gibbs free energy

R = Gas constant (0.008314 kJ mol^-1 K^-1)

T = Temperature

Put the values,

Keq = e^{-(-67.432kJ/mol/0.008314 kJ mol-1 K-1*298.15K)}

Keq = e^27.2033

Keq = 6.52 * 10⁸ [Answer]

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Answer -

Given,

Temperature = 298.15 K

Keq = ?

2 HBr (g)  $\rightarrow$ H₂ (g) + Br₂ (l)

Some Useful information,

Species	$\Delta$G$\degree$f (kJ/mol)
HBr (g)	-53.45
H2 (g)	0
Br2 (l)	0

We know that,

$\Delta$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 H₂ (g))) + 1 * $\Delta$ G^$\degree$_f Br₂ (l)) - 2 *$\Delta$G^$\degree$_f (HBr (g))

Put the values,

$\Delta$G^$\degree$_rxn = 1 * 0 kJ/mol) - 1 * 0 kJ/mol) - 2 * -53.45 kJ/mol)

$\Delta$G^$\degree$_rxn = 53.45 kJ /mol

We know that,

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

^{where, Keq = Equilibrium constant}

$\Delta$G^$\degree$ = Gibbs free energy

R = Gas constant (0.008314 kJ mol^-1 K^-1)

T = Temperature

Put the values,

Keq = e^{-(53.45kJ/mol/0.008314 kJ mol-1 K-1*298.15K)}

Keq = e-^21.56

Keq = 4.33 *10^-9 [Answer]

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