Reaction of water and methane
Polar and non-polar liquids do not mix, but that doesn't tell you anything about their ability to react, which depends on the thermodynamics and kinetics of the reaction. The reaction shown is in the gaseous state.

The reaction of methane gas with gaseous water, which you show, is a common reaction called Methane reformation. Methane gas is heated with steam (gaseous water) to produce hydrogen (and $\ce{CO2}$). 

As shown in that link, the reaction takes place in two steps: 
$$\begin{align}\ce{CH4_{(g)} + H2O_{(g)} &amp;-&gt;  CO_{(g)}  + H2_{(g)}} &amp;&amp;\text{(Endothermic)}\\
\ce{CO_{(g)} + H2O_{(g)} &amp;-&gt; CO2_{(g)} + H2_{(g)}} &amp;&amp; \text{(Exothermic)}\end{align}$$

It appears it probably requires some heating to get this reaction to proceed (the first step is endothermic), I'll edit when I have a chance to look up the thermodynamic properties and see if we can glean what drives the reaction, or if it requires catalysis to avoid the energy barrier.

Side-note: by "bipolar movement" I think you mean "dipole moment" 

Edit: According to this source, the first step of the reaction happens at ~3–20 atmospheres of pressure. The second step uses a catalyst to make the reaction more energetically favorable. 

The thermodynamic variables for the individual species (at $25~\mathrm{^\circ C}$) in the net reaction are:

$$
\begin{array}{c|cc}
 &amp; \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) &amp; \Delta H_\mathrm{f}^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right) \\
\hline
\ce{CH4_{(g)}} &amp; 0.1862 &amp; -74.81 \\ 
\ce{H2O_{(g)}} &amp; 0.1887 &amp; -241.8 \\
\ce{CO2_{(g)}} &amp; 0.2136 &amp; -393.5 \\
\ce{H2_{(g)}} &amp; 0.1306 &amp; 0 \\
\end{array}
$$ 

The balanced net reaction is:

$$ \ce{CH4_{(g)} + 2 H2O_{(g)} -&gt; CO2_{(g)} + 4 H2_{(g)}}$$

Therefore: 

$$
\begin{array}{l|ccc}
 &amp; \Delta S^\circ \left( \frac{\mathrm{kJ}}{\mathrm{K\cdot mol}} \right) &amp; \Delta H^\circ \left( \frac{\mathrm{kJ}}{\mathrm{mol}} \right)  \\
\hline
\text{Reactants} &amp; 0.5636 &amp; -558.41 \\ 
\text{Products} &amp; 0.7576 &amp; -393.5 \\
\hline
\text{Net Reaction} &amp; 0.2136 &amp; 164.9 \\
\end{array}
$$ 

Right away, it's apparent that the reaction is entropically favorable ($S_{\text{rxn}} &gt; 0$), but enthalpically unfavorable ($H_{\text{rxn}} &gt; 0$) at room temperature. To find out if the reaction is spontaneous, consider: 

$$
\Delta G = \Delta H -T\Delta S
$$

The reaction is spontaneous when $\Delta G &lt; 0$. Because $\Delta S$ is so close to $0$ ($0.194~\mathrm{kJ/(K\,mol)}$), and $\Delta H$ is so high ($164.91~\mathrm{kJ/mol}$), we should expect already that this reaction is non-spontaneous at room temperature. This is consistent with the fact that methane doesn't readily decompose in moist air, as any cow pasture will tell you.  But we can check:

$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-298.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=107.07\ \frac{\mathrm{kJ}}{\mathrm{mol}}$$

Quite non-spontaneous. However, since the overall reaction is entropically favorable, we can use $\Delta S$ to overpower the $\Delta H$, by jacking up $T$. Since the industrial reaction takes place between $700$ and $1000~\mathrm{^\circ C}$, let's plug in $700~\mathrm{^\circ C}\ (973.15~\mathrm{K})$ and check that the reaction is indeed spontaneous: 

$$\Delta G = 164.91\ \frac{\mathrm{kJ}}{\mathrm{mol}}-973.15~\mathrm{K}\cdot0.194~\frac{\mathrm{kJ}}{\mathrm{K\cdot mol}}=-23.85 \frac{\mathrm{kJ}}{\mathrm{mol}}$$

Voilà! A spontaneous reaction!

Do note, however, that this calculation:


ignores any temperature-dependence of the thermodynamic variables;
ignores the role of pressure. 


Tldr: This reaction is not energetically favorable at room temperature. However, the reaction is entropically favorable. If you heat it up enough (i.e. $\gtrapprox 700~\mathrm{^\circ C}$), the reaction will proceed spontaneously. 

Source for thermodynamic values: Oxtoby, Principles of Modern Chemistry Seventh Edition