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1 year ago

Liquid hydrogen peroxide, an oxidizing agent in many rocket fuel mixtures, releases oxygen gas on decomposition:

Chemistry

1 answer:

Elenna [48]1 year ago

4 0

The given reaction is as follows:

2H₂O₂(l) → 2H₂O(l) + O₂(g) ΔH = -196.1kj

The decomposition of 2 moles of hydrogen peroxide in the aforementioned reaction results in the emission of 196.1 kJ of heat.

1.88x106 kJ heat is released when 652 kg of H₂O₂ decomposes.

<h3>Solution ;</h3>

The formula to calculate moles is as follows:

Moles = Mass / Molar Mass

The molar mass of H2O2 is 34.01 g/mol.

On substituting values in formula,

Moles = 361kg

34.01g / mol =361000g / 34.01g/mol (Since,1kg=1000g) = 10,614.53mol

From the stoichiometry, 2 moles of hydrogen peroxide release 196.1 kJ of heat.

So, 10,614.53 moles of hydrogen peroxide will release 10,614.532×196.1=1.04×10^6kJ of heat.

So finally we can say that 1.88x106 kJ heat is released when 652 kg of H₂O₂ decomposes.

To know more about Liquid Hydrogen Peroxide please click here : brainly.com/question/1459322

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Answer:

The equilibrium constant is $K =0.02867$

The equilibrium pressure for oxygen gas is $P_{O_2} = 0.09367\ atm$

Explanation:

From the question we are told that

The equation of the chemical reaction is

$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

The Gibbs free energy for $M_3 O_4_{(s)}$ is $\Delta G^o_{1} = -8.80 \ kJ/mol$

The Gibbs free energy for $M{(s)}$ is $\Delta G^o_{2} = 0 \ kJ/mol$

The Gibbs free energy for $O_2{(s)}$ is $\Delta G^o_{3} = 0 \ kJ/mol$

The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$