1
Question
Heat of formation of CH4 is:
If given:
C(s)+O2(g)→CO2(g) △H=−394 kJ
H2(g)+12O2(g)→H2O(l) △H=−284 kJ
CH4(g)+2O2(g)→CO2(g)+2H2O(l) △H=−892 kJ
Heat of formation of CH4 is:
If given:
C(s)+O2(g)→CO2(g) △H=−394 kJ
H2(g)+12O2(g)→H2O(l) △H=−284 kJ
CH4(g)+2O2(g)→CO2(g)+2H2O(l) △H=−892 kJ
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Solution
The correct options are
A -70 kJ
B -16.7 kcal
(A,B)
C(s)+O2(g)→CO2(g) △H=−394 kJ .......(i)
H2(g)+12O2(g)→H2O(l) △H=−284 kJ.......(ii)
CH4(g)+2O2(g)→CO2(g)+2H2O(l) △H=−892 kJ...........(iii)
Adding (i) & (2 ×(ii)) and subtracting (iii) from it,
-394 + (2 ×(−284)) + 892
=-70 kJ
1 calorie = 4.18 J
⇒−70kJ=−16.7kcal.
A
-70 kJ
B
-16.7 kcal
(A,B)
C(s)+O2(g)→CO2(g) △H=−394 kJ .......(i)
H2(g)+12O2(g)→H2O(l) △H=−284 kJ.......(ii)
CH4(g)+2O2(g)→CO2(g)+2H2O(l) △H=−892 kJ...........(iii)
Adding (i) & (2 ×(ii)) and subtracting (iii) from it,
-394 + (2 ×(−284)) + 892
=-70 kJ
1 calorie = 4.18 J
⇒−70kJ=−16.7kcal.
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