According to the following thermodynamic data, calculate the enthalpy $(10\%),$ entropy $(10\%)$ and Gibbs free energy $(10\%)$ of the reaction $Pb+\frac{1}{2}{O}_2=$PbO at $1000K.$

$H_{\text{P}\text{b}\text{O}}(298)=-219,000\frac{J}{m}$ole

$S_{Pb(298)}=65\frac{J}{K}$ mole

SPbO$(298)=66\mathrm{.}3\frac{J}{K}$ mole

$S_{O2(298)}=205\frac{J}{K}$ mole

$C_p,Pb(s)=23\mathrm{.}6+{9\mathrm{.}75}^{**}{10}^{-3}T\frac{J}{K}$ from $298K$ to $T_m*Pb$

$C_p,Pb(l)=32\mathrm{.}4-3\mathrm{.}1**{10}^{-3}T\frac{J}{K}$ from $T_{m*Pb}$ to $1200K$

$C_{p,\text{P}\text{b}\text{O}(s)}=37\mathrm{.}9+{26\mathrm{.}8}^{**}{10}^{-3}T\frac{J}{K}$ from $298K$ to $T_{m*PbO}$

$C_p,O2(g)=29\mathrm{.}96+{4\mathrm{.}18}^{**}{10}^{-3}T-1\mathrm{.}67**{10}^5{T}^{-2}\frac{J}{K}$ from $298K$ to $3000K$

$H_{m,Pb}=4810J$ at $T_{m,Pb}=600K$

$T_{m,\text{P}\text{b}\text{O}}=1159K$