Question

when NH4NO2 decomposes at 373 K it forms N2 and H2O. the ∆H for the reaction at 1 atmosphere pressure at 373 K is -223.6 KJ/mol .What is the value of ∆U for the reaction​

Answer 1

Explanation:

∆u= q+ w

q = -223.6

w = ∆ngRT

∆n = 2

R = 8.303

T = 373 put and get answer

Answer 2

The value of ∆U for the reaction​ is -232.9kJ/mol

Explanation:

The balanced decomposition reaction is shown by:

$NH_4NO_2(s)\rightarrow N_2(g)+2H_2O(g)$

Formula used :

$\Delta H=\Delta U+\Delta n_g\times RT$

where,

$\Delta H$ =  enthalpy of the reaction= -223.6 kJ/mol = -22360 J/mol

$\Delta U$= internal energy of the reaction = ?

$\Delta n_g$ = change in the moles of the reaction = Moles of product - Moles of reactant = 3 - 0 = 3 mole

R = gas constant = 8.314 J/mole.K

T = temperature = 373 K

Putting in the values we get:

$-22360=\Delta U+3\times 8.314\times 373$

$\Delta U=-232903J/mol=-232.9kJ/mol$

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