# Enthalpy chemistry calculation help

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#1
**** Use ΔHf data from the table to calculate a value for the enthalpy change for the following reaction.
NH4NO3(s)* +* C(s)* →* N2(g)* +* 2H2O(g)* +* CO2(g)

*
Substance
C(s)
N2(g)
H2O(g)
CO2(g)
NH4NO3(s)
ΔHf/ kJ mol–1
0
0
–242
–394
–365

(values given respectively)
I got -513 kj/mol but the answer is -316 kj/mol.
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1 year ago
#2
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1 year ago
#3
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1 year ago
#4
You have more oxygen atoms on the products than on the reactants. To balance the equation just put 1/2 mole for carbon and CO2.
The enthalpy change on reactants: -365 and on products: 0 + 2(-242) + 1/2(-394) = -618
∆Hr = ∆Hf2 -∆Hf1
= -618 -( -365 )
= -316 kJmol -1
0
1 year ago
#5
(Original post by nia(the_devoted))
You have more oxygen atoms on the products than on the reactants. To balance the equation just put 1/2 mole for carbon and CO2.
The enthalpy change on reactants: -365 and on products: 0 + 2(-242) + 1/2(-394) = -618
∆Hr = ∆Hf2 -∆Hf1
= -618 -( -365 )
= -316 kJmol -1
-618 -(-365) -316 = -253
It's a typo, you meant to write -681 kJmol^-1

balanced eqn
NH4NO3(s) + 1/2C(s) = N2(g) + 1/2 CO2(g) + 2H2O(g)
∆H = ∆H(products) - ∆H(reactants) = 1(-242) + 1/2(-394) - (-365) = -316 kJmol^-1
0
1 year ago
#6
(Original post by BobbJo)
-618 -(-365) -316 = -253
It's a typo, you meant to write -681 kJmol^-1

balanced eqn
NH4NO3(s) + 1/2C(s) = N2(g) + 1/2 CO2(g) + 2H2O(g)
∆H = ∆H(products) - ∆H(reactants) = 1(-242) + 1/2(-394) - (-365) = -316 kJmol^-1
yeah, thanks.
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