Answer
$9.1\,kJ$
The reaction is non-spontaneous.
Work Step by Step
$\Delta H^{\circ}=\Sigma n_{p}\Delta H_{f}^{\circ}(products)-\Sigma n_{r}\Delta H_{f}^{\circ}(reactants)$
$=[2\Delta H_{f}^{\circ}(NO_{2},g)]-[2\Delta H_{f}^{\circ}(NO,g)+\Delta H_{f}^{\circ}(O_{2},g)]$
$=[2(33.2\,kJ/mol)]-[2(91.3\,kJ/mol)+(0)]$
$=-116.2\,kJ$
$\Delta S^{\circ}=\Sigma n_{p}S^{\circ}(products)-\Sigma n_{r}S^{\circ}(reactants)$
$=[2S^{\circ}(NO_{2},g)]-[2S^{\circ}(NO,g)+S^{\circ}(O_{2},g)]$
$=[2(240.1\,Jmol^{-1}K^{-1})]-[2(210.8\,Jmol^{-1}K^{-1})+(205.2\,Jmol^{-1}K^{-1})]$
$=-146.6\,JK^{-1}=-0.1466\,kJK^{-1}$
$\Delta G=\Delta H-T\Delta S$
$=(-116.2\,kJ)-(855\,K)(-0.1466\,kJK^{-1})$
$=9.1\,kJ$
Because $\Delta G$ is positive, the reaction is non-spontaneous.
