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

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

Step-by-step 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}`