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Snowcat [4.5K]
3 years ago
9

Which of the following has the correct change in enthalpy and reaction description for the following reaction? NH3(g) + HCl(g) N

H4Cl(s) Given: NH3: ∆H= -46.2 kJ HCl: ∆H= -92.3 kJ NH4Cl: ∆H= -314.43 kJ
Chemistry
2 answers:
Molodets [167]3 years ago
4 0
 <span>2HCl -----> H2 + Cl2 dH = +184.6kJ/mol (equation 1 reversed) 
2NH3 ------> 3H2 + N2 dH = +92.20kJ/mol ( equation 2 unchanged) 
N2 + 4H2 + Cl2 ------> 2NH4Cl dH = - 628.8kJ/mol (equation 3 doubled) 
--------------------------------------... 
Adding up: 2HCl + 2NH3 -----> 2NHCl4 dH = -352kJ/mol 
Hence HCl + NH3 --------> NH4Cl dH = -352/2 = - 176kJ/mol.</span>
Usimov [2.4K]3 years ago
4 0
To find the change in enthalpy (ΔH), all you have to do is use the following formula:

ΔH reaction= [ΔH of products] - [ΔH of reactants]

remember that the products are the ones in the right side of the reaction, and reactants are in the left side. 

ΔH reaction= [ΔH NH4Cl] - [ΔH NH3 + ΔH HCl]

ΔH reaction= [-314.43] - [ -46.2 + -92.3]= -175.93 kJ
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Explanation:

To solve this kind of exercises you must look for the number of atoms in each molecule first, then look on the periodic table the atom weight and the multiply the atom weight times the quantity of each atom. For instance:

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See you,

5 0
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