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-BARSIC- [3]
4 years ago
5

Name two methods that can be used to break down compounds into simpler substance

Chemistry
2 answers:
Nezavi [6.7K]4 years ago
8 0
<span>The two methods that can be used do break down compounds into simpler substances or smaller units, are known as Hydrolysis and Photolysis. 

Hydrolysis uses water to split a molecule. Water is used to break the bonds of a molecule. There are three main types of hydrolysis:</span>1. acid hydrolysis2. base hydrolysis 3. salt hydrolysis

Photolysis uses energy from light to split molecules.  This energy is called photons. As in hydrolysis, photons are used to break the bonds of a molecule. Photolysis is a chemical reaction. 
kotykmax [81]4 years ago
7 0
<span>Photolysis and hydrolysis. These are two methods that can be used to break down a compound into simpler substances and smaller units.

Water which is used to break the bonds of molecules and split molecules is used from hydrolysis. Hydrolysis is made of three types which include;
1. salt hydrolysis.
2. acid hydrolysis.
3. Base hydrolysis.
 Photolysis is well known to use energy from light to split the molecule and the same energy is referred to as photons which are used to break builds of molecules.</span>
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<u>Answer:</u> The standard heat for the given reaction is -138.82 kJ

<u>Explanation:</u>

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

For the given chemical reaction:

4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]

We are given:

\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol

Putting values in above equation, we get:

\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ

Hence, the standard heat for the given reaction is -138.82 kJ

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