Question

Consider the reaction 2N2(g) O2(g)2N2O(g) Using the standard thermodynamic data in the tables linked above, calculate Grxn for this reaction at 298.15K if the pressure of each gas is 22.20 mm Hg.

Answer 1

`\Delta G^0 _(rxn) = 207.6\ kJ/mol`

ΔG ≅ 199.91 kJ

Step-by-step explanation:

Consider the reaction:

`2N_(2(g)) + O_(2(g)) \to 2N_2O_((g))`

temperature = 298.15K

pressure = 22.20 mmHg

From, The standard Thermodynamic Tables; the following data were obtained

`\Delta G_f^0 \ \ \ N_2O_((g)) = 103 .8 \ kJ/mol`

`\Delta G_f^0 \ \ \ N_2{(g)} =0 \ kJ/mol`

`\Delta G_f^0 \ \ \ O_2{(g)} =0 \ kJ/mol`

`\Delta G^0 _(rxn) = 2 * \Delta G_f^0 \ N_2O_((g)) - ( 2 * \Delta G_f^0 \ N_2{(g)} + \Delta G_f^0 \ O_(2(g)))`

`\Delta G^0 _(rxn) = 2 * 103.8 \ kJ/mol - ( 2 * 0 + 0)`

`\Delta G^0 _(rxn) = 207.6\ kJ/mol`

The equilibrium constant determined from the partial pressure denoted as
`K_p` can be expressed as :

`K_p = ((22.20)^2)/((22.20)^2 * (22.20))`

`K_p = (1)/( (22.20))`

`K_p` = 0.045

`\Delta G = \Delta G^0 _(rxn) + RT \ lnK`

where;

R = gas constant = 8.314 × 10⁻³ kJ

`\Delta G =207.6 + 8.314 * 10 ^(-3) * 298.15 \ ln(0.045)`

`\Delta G =207.6 + 2.4788191 * \ ln(0.045)`

`\Delta G =207.6+ (-7.687048037)`

`\Delta G =` 199.912952 kJ

ΔG ≅ 199.91 kJ