How much heat is evolved when 12 g
of CO reacts with NO, ? The reaction
is :
4 CO(g) + 2 NO2(g) →
4 CO (9) +N (9), ∆H' = -1200 kJ
Answers
Answer:I Im taking 14kJ But Procedure is same...
Explanation: Hope it will help you out...
The problem provides you with the thermochemical equation for the combustion of carbon monoxide,
CO
2
CO
2(g]
+
O
2(g]
→
2
CO
2(g]
Δ
H
rxn
=
−
560 kJ
In essence, the thermochemical equation tells you the enthalpy change of reaction,
Δ
H
rxn
, for when the reaction consumes the number of moles of reactants and produces the number of moles of product(s) that act as stoichiometric coefficients in the balanced chemical equation.
In this case, you can say that when
2
moles of carbon monoxide react with
1
mole of oxygen gas,
2
moles of carbon dioxide are produced and the enthalpy change of reaction is
Δ
H
rxn
=
−
560 kJ
.
In other words, when
2
moles of carbon monoxide undergo combustion,
560 kJ
" of heat are being given off, hence the minus sign used for
Δ
H
rxn
.
So, you know that the reaction gives off
560 kJ
of heat when two moles of carbon monoxide react. Use the compound's molar mass to determine how many moles you get in that
14.0-g
sample
14.0
g
⋅
molar mass of CO
1 mole CO
28.0101
g
=
0.4998 moles CO
You can now use the known enthalpy change of reaction for when two moles of carbon monoxide undergo combustion as a conversion factor to figure out how much heat is being given off when
0.4998
moles react
0.4998
moles CO
⋅
the known
a
Δ
H
rxn
560 kJ
2
moles CO
=
139.94 kJ
Rounded to two sig figs, the number of sig figs you have for
Δ
H
rxn
, the answer will be
heat given off
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
140 kJ
a
a
∣
∣
−−−−−−−−−−
This is equivalent to saying that the enthalpy change of reaction when
14.0 g
of carbon monoxide undergo cumbustion is equal to
Δ
H
rxn for 14.0 g
=
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
−
140 kJ
a
a
∣
∣
−−−−−−−−−−−−−
Remember, the minus sign denotes heat given off.