Question

calculate $\delta_rg^\theta_m$, judge the spontaneity.
$2ch_3oh(l) \
ightarrow 2ch_4(g) + o_2(g)$
$\delta_fh^\theta_m/(kj\cdot mol^{-1})$ $s^\theta_m/(j\cdot mol^{-1}\cdot k^{-1})$
$ch_3oh(l)$: -239.2 126.8
$ch_4(g)$: -74.6 186.3
$o_2(g)$: 0 205.2
do by yourself

Explanation:

Step1: Calculate $\Delta_{r}H_{m}^{\ominus}$

The formula for enthalpy change of reaction is $\Delta_{r}H_{m}^{\ominus}=\sum
u\Delta_{f}H_{m}^{\ominus}(\text{products})-\sum
u\Delta_{f}H_{m}^{\ominus}(\text{reactants})$.
For the reaction $\ce{2CH_{3}OH(l) -> 2CH_{4}(g) + O_{2}(g)}$:
$\Delta_{f}H_{m}^{\ominus}(\ce{CH_{3}OH(l)})=-239.2\ \text{kJ·mol}^{-1}$, $\Delta_{f}H_{m}^{\ominus}(\ce{CH_{4}(g)})=-74.6\ \text{kJ·mol}^{-1}$, $\Delta_{f}H_{m}^{\ominus}(\ce{O_{2}(g)}) = 0\ \text{kJ·mol}^{-1}$.
$\sum
u\Delta_{f}H_{m}^{\ominus}(\text{products})=2\times(-74.6)+1\times0=-149.2\ \text{kJ·mol}^{-1}$
$\sum
u\Delta_{f}H_{m}^{\ominus}(\text{reactants})=2\times(-239.2)= - 478.4\ \text{kJ·mol}^{-1}$
$\Delta_{r}H_{m}^{\ominus}=-149.2-(-478.4)=329.2\ \text{kJ·mol}^{-1}$

Step2: Calculate $\Delta_{r}S_{m}^{\ominus}$

The formula for entropy change of reaction is $\Delta_{r}S_{m}^{\ominus}=\sum
u S_{m}^{\ominus}(\text{products})-\sum
u S_{m}^{\ominus}(\text{reactants})$.
$S_{m}^{\ominus}(\ce{CH_{3}OH(l)}) = 126.8\ \text{J·mol}^{-1}·\text{K}^{-1}$, $S_{m}^{\ominus}(\ce{CH_{4}(g)})=186.3\ \text{J·mol}^{-1}·\text{K}^{-1}$, $S_{m}^{\ominus}(\ce{O_{2}(g)}) = 205.2\ \text{J·mol}^{-1}·\text{K}^{-1}$.
$\sum
u S_{m}^{\ominus}(\text{products})=2\times186.3 + 1\times205.2=372.6 + 205.2 = 577.8\ \text{J·mol}^{-1}·\text{K}^{-1}$
$\sum
u S_{m}^{\ominus}(\text{reactants})=2\times126.8 = 253.6\ \text{J·mol}^{-1}·\text{K}^{-1}$
$\Delta_{r}S_{m}^{\ominus}=577.8 - 253.6 = 324.2\ \text{J·mol}^{-1}·\text{K}^{-1}=0.3242\ \text{kJ·mol}^{-1}·\text{K}^{-1}$

Step3: Calculate $\Delta_{r}G_{m}^{\ominus}$ at 298K (assuming standard temperature)

The formula is $\Delta_{r}G_{m}^{\ominus}=\Delta_{r}H_{m}^{\ominus}-T\Delta_{r}S_{m}^{\ominus}$, $T = 298\ \text{K}$.
$\Delta_{r}G_{m}^{\ominus}=329.2-298\times0.3242$
$=329.2 - 96.6116=232.5884\ \text{kJ·mol}^{-1}$

Answer:

$\Delta_{r}G_{m}^{\ominus}\approx232.6\ \text{kJ·mol}^{-1}$ (Since $\Delta_{r}G_{m}^{\ominus}>0$, the reaction is non - spontaneous at 298K and standard conditions.)