Question

Substance AGP (kJ/mol) AS (J/mol K) NO (g) 86.7 211 NO2 (g) 51.8 240 NOCI (g) 66.3 264 N20 (g) 103.6 220 What is the value of AH° (in kJ) at 25.0 °C for the reaction: N20 (g) + NO2 (g) → 3 NO (g) 5.5 kJ 53.2 kJ 109 kJ 156.2 kJ

Answer 1

Given the reaction:

N2O(g) + NO2(g) + 3NO(g)

From thermodynamics, we know that,

AG AH TAS

Therefore,

$\small \Delta G^o +T\Delta S^o= \Delta H^o$

or,

\H° = {G° +TAS° ... (1)

i) Calculation of :

10 Gº

$\small \Delta G^o = \sum_{products}m G^o_{products} -\sum_{reactants}n G^o_{reactants}$

where m and n are the stoichiometric coefficients of products and reactants, respectively

Therefore, for the given reaction we can write,

$\small \Delta G^o = 3G^o_{NO} - \left (G^o_{NO_2}+G^o_{N_2O} \right )$

$\small \Delta G^o = (3mol)(86.7kJ/mol) - \left ( (1mol)(51.8kJ/mol)+(1mol)(103.6kJ/mol) \right )$

AG° = (3 mol )(86.7kJ/mol )-((1 md)(51.8kJ/mol ) + (1 pro)(103.6kJ/mol

$\small \Delta G^o = (3)(86.7kJ) - \left ( (1)(51.8kJ)+(1)(103.6kJ) \right )$

$\small \Delta G^o = 260.1kJ - (51.8kJ)-(103.6kJ)$

$\small \Delta G^o = 104.7\;kJ$

ii) Calculation of $\small \Delta S^o$

$\small \Delta S^o = \sum_{products}m S^o_{products} -\sum_{reactants}n S^o_{reactants}$

where m and n are the stoichiometric coefficients of products and reactants, respectively

:Therefore, for the given reaction we can write,

$\small \Delta S^o = 3S^o_{NO} - \left (S^o_{NO_2}+S^o_{N_2O} \right )$

$\small \Delta S^o = (3mol)(211J/molK) - \left ( (1mol)(240J/molK)+(1mol)(220J/molK) \right )$

$\small \Delta S^o = (3\cancel{mol})(211J/\cancel{mol}K) - \left ( (1\cancel{mol})(240J/\cancel{mol}K)+(1\cancel{mol})(220J/\cancel{mol}K) \right )$

$\small \Delta S^o = (3)(211J/K) - (240J/K)-(220J/K) \right )$

$\small \Delta S^o =173 \;J/K$

or,

(1 J = 1/1000 kJ)

$\small \Delta S^o =0.173 \;kJ/K$

iii) Calculation of $\small \Delta H^o$

Given temperature (T) = 25$\small ^oC$

T(in Kelvin) = 273 + T (in $\small ^oC$ )

T(in Kelvin) = 273 + 25

T(in Kelvin) = 298 K

Now, using (1), that is

$\small \Delta H^o =\Delta G^o +T\Delta S^o\;$

$\small \Delta H^o =104.7kJ +(298K)(0.173\;kJ/K)\;$

$\small \Delta H^o =104.7kJ +(298)(0.173\;kJ)\;$

$\small \Delta H^o =104.7kJ +51.5kJ$

$\small \Delta H^o =156.2 kJ$

Hence, the correct answer is $\small \Delta H^o =156.2 kJ$