Hydrogen Peroxide and Potassium Permanganate as Combustionless Fuel

*All constants used for this project, unless otherwise stated, were found in the 97th edition of the CRC Handbook of Chemistry and Physics

In searching for a combustionless fuel, to be used in a rocket, we will be concerning ourselves with the reaction

$3H_{2}O_{2}(aq) + 2KMnO_{4}(aq) \longrightarrow 2MnO_{2}(c)+2H_{2}O(l)+3O_{2}(g)+2KOH(aq)$ .

It is known that hydrogen peroxide spontaneously decomposes into water and oxygen via the reaction

$2H_{2}O_{2} \longrightarrow 2H_{2}O+O_{2}$ .

Moreover, it is well established that potassium permanganate accelerates the decomposition of hydrogen peroxide via the reaction

$3H_{2}O_{2}(aq) + 2KMnO_{4}(aq) \longrightarrow 2MnO_{2}(c)+2H_{2}O(l)+3O_{2}(g)+2KOH(aq)$ .

In this part of our project, we wish to establish the enthalpy of a complete reaction as well as the rate of reaction as a function of the concentration of our reactants in an aqueous form. The reason for this is that both the enthalpy of a complete reaction and the rate of reaction will allow us to calculate theoretical values for the peak performance of the said rocket.

The following post will document our work in these endeavors.

Enthalpy:

As the CRC Handbook fails to provide the enthalpy of formation for $H_{2}O_{2}(aq)$ and $KMnO_{4}(aq)$ we will for now calculate the enthalpy of our reaction the enthalpy of $H_{2}O_{2}(l)$ and $KMnO_{4}(c)$ . We will later compare this to data about the heat released by our reaction which will be collected while trying to establish the rate of reaction as a function of the concentration of $KMnO_{4}(aq)$ . We anticipate one of four outcomes will be seen in those data:

Enthalpy of our reaction is constant with small percent error when compared to our theoretical calculations
- Conclusion:
  our assumption that we could use the enthalpy of $H_{2}O_{2}(l)$ and $KMnO_{4}(c)$ to calculate the heat released by our reaction was correct
Enthalpy of our reaction is constant with large percent error when compared to our theoretical calculations
- Conclusion:
  the theoretical calculations should be made using values for $H_{2}O_{2}(aq)$ and $KMnO_{4}(aq)$
Heat released by our reaction varies with the concentration
- Conclusion:
  Enthalpy varies as a function of the concentration of $KMnO_{4}(aq)$ and more work is needed to establish relations for both $H_{2}O_{2}(aq)$ and $KMnO_{4}(aq)$ . In that case, our experiments will be followed up by experimentally finding the heat released by our reaction when four different concentrations of $H_{2}O_{2}(aq)$ are added to four different concentrations of $KMnO_{4}(aq)$ . It is our belief that the resulting 16 points of data will be sufficient to establish a relationship between the heat released by our reaction and the concentrations of both $H_{2}O_{2}(aq)$ and $KMnO_{4}(aq)$ .
Heat released by our reaction varies, but with no clear dependence on the concentration of .
- Conclusion:
  Colaboration with a profesor or graduate student in the Chemistry dept. is necessary to proceed.

Calculations:

n\left(\Delta H^{\circ}(Y)\right)_{n[\frac{kJ}{mol}]}

Hence the enthalpy of formation for this particular reaction is:
$\sum\Delta H^{\circ}(Y)-\sum\Delta H^{\circ}(X)=-223.00_{kJ}$ per each complete reaction.

Given that the molar mass of $KMnO_{4}$ is $158.03_{\frac{g}{mol}}$ and the molar mass of $H_{2}O_{2}$ is $34.01_{\frac{g}{mol}}$ that means the heat density of our reaction per $g$ of $KMnO_{4}$ is $-0.71_{\frac{kJ}{g}}$ and per $g$ of $H_{2}O_{2}$ is $-2.19_{\frac{kJ}{g}}$ .

Quick Napkin Math:

That means if we assume the complete reaction of $5_{kg}$ of a $20\%$ concentrated solution of $H_{2}O_{2}$ with a solution of $KMnO_{4}$ then we will produce $-2,190_{kJ}$ of heat.

Now, assuming we wish to lift a mass of $136_{kg}$ a distance of $30_{m}$ , we need to expel $39,984_{J}$ or $39.98_{kJ}$ . That means we only need to convert $2\%$ of the total heat to kinetic energy!

…yea. I can work with that. Bet!

Products(nY)	$n\left(\Delta H^{\circ}(Y)\right)_{n[\frac{kJ}{mol}]}$	$\sum \left(\Delta H^{\circ}(Y)\right)$
$3O_{2}(g)$	$3(0.00)$	$0.00$
$2MnO_{2}(c)$	$2(-520.00)$	$-1,040.00$
$2KOH(c)$	$2(-424.60)$	$-849.20$
$2H_{2}O(l)$	$2(-285.8)$	$-571.60$
$\sum n\left(\Delta H^{\circ}(Y)\right)$	–	$-2,460.80$

Reactants(X)	$n\left(\Delta H^{\circ}(X)\right)$	$\sum \left(\Delta H^{\circ}(X)\right)$
$2KMnO_{4}$	$2(-837.2)$	$-1,674.40$
$3H_{2}O_{2}$	$3(-187.80)$	$-563.40$
$\sum n\left(\Delta H^{\circ}(X)\right)$	–	$-2,237.80$

Enthalpy:

Calculations:

Quick Napkin Math:

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