Question

The value of ΔG° (kJ/mol) at 25°C for the formation of phosphorous trichloride from its constituent elements: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) A) –59.2 B) +59.2 C) –642 D) +75.6 E) –373 F) –29.4

Answer 1

Answer:

F.    $\mathsf{\Delta G^0 \simeq -29.4 \ kJ}$

Explanation:

GIven that:

The chemical equation for the reaction is:

$\mathtt{Fe_2O_{3(s)} + 3CO_{(g)} \to 2Fe _{(s)} + 3CO_{2(g)}}$

The value of the ΔG° (kJ/mol) at 25 °C

$\Delta G^0 = \begin {pmatrix} 2( \Delta G_f^0 \ Fe_{(s)} )+ 3 ( \Delta G_f^0 \ CO_{2(g)} ) - 1 ( \Delta G_f^0 \ Fe_2CO_3_{(s)} + 3 ( ( \Delta G_f^0 \ CO _{(g)} ) \end {pmatrix}$

where;

$\Delta G_f^0 \ Fe_{(s)} = 0$

$\Delta G_f^0 \ CO_{2(g)} = -394.359 \ kJ/mole$

$\Delta G_f^0 \ Fe_2CO_3_{(s)} = -742.2 \ kJ/mole$

$\Delta G_f^0 \ CO _{(g)}= -137.168\ kJ/mole$

Replacing the values ; we have

$\Delta G^0 = \begin {pmatrix} 2(0 )+ 3 ( -394.359 ) ) -( 1 (-742.2) + 3 (-137.168 ) \end {pmatrix}$

$\Delta G^0 = \begin {pmatrix} 0 -1183.077 ) -(-742.2- 411.504 \end {pmatrix}$

$\Delta G^0 = \begin {pmatrix} -1183.077 +742.2+ 411.504 \end {pmatrix}$

$\mathtt{\Delta G^0 = -29.373 \ kJ}$

$\mathsf{\Delta G^0 \simeq -29.4 \ kJ}$