Combustion Jet

The combustion of methane proceeds according to $\ce{CH4 + 2O2 -> CO2 + 2H2O,}$ forming one molecule of carbon dioxide and two molecules of water for each molecule of methane combusted. If we assume that $100\%$ of the energy that's released goes into the kinetic energy of the products, then we have $E_\textrm{total} = \frac12 m_{\ce{CO2}} u_{\ce{CO2}}^2 + 2\times \frac12 m_{\ce{H2O}}u_{\ce{H2O}}^2.$ Now, the atomic mass of carbon dioxide and water are a bit different ( $44$ and $18,$ respectively) but by multiplying and dividing the right hand side by the sum of the masses, we can find a characteristic value for the speed of a given exhaust molecule. $\begin{aligned} E_\textrm{total} &= \frac12 \frac{m_{\ce{CO2}} u_{\ce{CO2}}^2 + 2\times m_{\ce{H2O}}u_{\ce{H2O}}^2}{m_{\ce{CO2}} + 2m_{\ce{H2O}}}\left(m_{\ce{CO2}} + 2m_{\ce{H2O}}\right) \\ 2E_\textrm{total} &= \langle u_\textrm{e}^2\rangle\left(m_{\ce{CO2}} + 2m_{\ce{H2O}}\right) \end{aligned}$ where we've defined $\langle u_\textrm{e}^2\rangle$ as the mass-weighted mean square velocity of the exhaust molecules. Taking the root, we have an estimate for the maximum exhaust speed for the combustion of methalox (methane + liquid oxygen):

$\begin{aligned} u_\textrm{e} &\approx \sqrt{\frac{2E_\textrm{total}}{m_{\ce{CO2}} + 2m_{\ce{H2O}}}} \\ &\approx \sqrt{2\times\SI{890000}{\joule\per\mole}/\SI{0.08}{\kilo\gram\per\mole}} \\ &< \sqrt{20\times\frac{5}{4}\times\SI{1000000}{\joule\per\kilo\gram}} \\ &= \sqrt{25\times\SI{1000000}{\joule\per\kilo\gram}} \\ &= \SI{5000}{\meter\per\second}. \end{aligned}$

In practice, $u_e$ is something less than $\SI{5}{\kilo\meter\per\second},$ topping out near $\SI{3.6}{\kilo\meter\per\second}$ in the vacuum of space, and a bit less on Earth. But this gets us in the ballpark.

I solved it by assuming that $u_{CO_2}=u_{H_2O}$ , to get 4717m/s.

The exhaust gas is an ionized plasma, and there is a strong interaction between the particles. (We know that from the fact that the exhaust glows.) If there is a strong interaction between the particles, they will come to thermal equilibrium. The actual velocity of a given particle is $u=u_{thermal}+u_{average}$ , where the average is the same for all particles and the thermal velocity depends on the mass, calculated from the equipartition theorem, $u_{thermal}=\sqrt{\frac{3k_B T}{m}}$ , where $k_B$ is the Boltzmann constant, $T$ is the temperature and $m$ is the mass of the molecule. The number quoted above is valid only if we neglect the energy carried away by thermal motion. Taking into account the disordered motion in the hot gas (and the energy carried by the excited state of the ions) will make the actual exhaust velocity smaller than the estimated value, bringing it closer to the 3.6km/s value quoted by Josh.

I didn't know many about this type of reactions so what I did was:

Atomic mas for Oxigen: 16.

Atomic mas for Carbon: 12.

Atomic mas for Hidrogen: 1.

Atomic mas for all the reactives or products: 16 2+12 1+8*1=52.

Then with the converssion 52g/mol:

890kJ/mol (1mol/52g) (1000g/1kg)*(1000J/1kJ) to get the units J/kg=(m/s)^2. Then I suposed the result would be u^2 so I took the square root and got the correct answer with u=4137.07...m/s~5000m/s.