Question

Calculate deltaGϴ for the following balanced reaction at 25 °C using the data in the table below.

Calculate AG for the following balanced reaction at 25 °C using the data in the table below. CO(NH2)2(s) + H2O(l) → CO2(g) + 2 NH3(g) A:HⓇ (kJ mor') -333.19 Se (J K-'mol-') 104.6 CO(NH2)2(s) H2O(1) -285.9 69.96 CO2(g) -393.5 213.6 NH3(g) -46.19 192.5

Answer 1

Answer -

Given,

Temperature = 25^$\degree$ C or 298.15 K [0°C + 273.15 = 273.15 K]

	$\Delta$H$\degree$ (kJmol-1)	S$\degree$ (JK-1mol-1)
CO(NH2)2(s)	-333.19	104.6
H2O (l)	-285.9	69.96
CO2 (g)	-393.5	213.6
NH3 (g)	-46.19	192.5

CO(NH₂)₂ (s)  + H₂O (l) $\rightarrow$ CO₂ (g) + 2 NH3 (g) [BALANCED]

We know that,

$\Delta$H^$\degree$_rxn = $\sum$ n $\Delta$H^$\degree$(products) -  $\sum$ n $\Delta$H^$\degree$(reactants)

Also, $\Delta$S^$\degree$_rxn = $\sum$ n $\Delta$S^$\degree$(products) -  $\sum$ n $\Delta$S^$\degree$(reactants)

where, n = stiochiometric coefficients

So,

$\Delta$H^$\degree$_rxn = 1 mol * $\Delta$H^$\degree$(CO₂(g)) + 2 mol * $\Delta$H^$\degree$(NH₃(g)) - 1 mol $\Delta$H^$\degree$ (CO(NH₂)₂(s)) - 1 mol $\Delta$H^$\degree$ (H₂O (l))

Put the values,

$\Delta$H^$\degree$_rxn = (1 mol * -393.5 kJmol^-1) + (2 mol * -46.19 kJmol^-1) + (-1 mol* -333.19 kJmol^-1) +(-1 mol -285.9 kJmol^-1)

$\Delta$H^$\degree$_rxn = 133.21 kJ -----------------------A

1 kJ = 1000 J

So, 133.21 kJ = 133210 J

And,

$\Delta$S^$\degree$_rxn = 1 mol * $\Delta$S^$\degree$(CO₂(g)) + 2 mol * $\Delta$S^$\degree$(NH₃(g)) - 1 mol $\Delta$S^$\degree$ (CO(NH₂)₂(s)) - 1 mol $\Delta$S^$\degree$ (H₂O (l))

Put the values,

$\Delta$S^$\degree$_rxn = (1 mol * 213.6 JK^-1mol^-1) + (2 mol * 192.5 JK^-1mol^-1) - (1 mol* 104.6 JK^-1mol^-1) - (1 mol 69.96 JK^-1mol^-1)

$\Delta$S^$\degree$_rxn = 424.04 JK^-1 -----------------------B

Also,

$\Delta$G^$\degree$_rxn = $\Delta$H^$\degree$_rxn - (T *$\Delta$S^$\degree$_rxn)

Put the values,

$\Delta$G^$\degree$_rxn = 133210 J - (298.15 K *424.04 JK^-1)

$\Delta$G^$\degree$_rxn = 6782.474 J or 6.78 kJ [Answer]