wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Given, C(graphites)+O2(g)CO2(g)
Δr H=393.5 kJ mol1
H2(g)+12 O2(g)H2O(l);
ΔrH=285.8 kJ mol2
CO2(g+2H2O(l)CH4(g)+2O2(g)
ΔrH=+890.3 kJ mol1
Based on the above thermochemical equations, the value of Δr H at 298K for the reaction, C(graphite)+2H2(g)CH4(g) will be

A
+78.8kJmol(1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
+144.0kJmol(1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
74.8kJmol(1)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
144.0kJmol(1)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 74.8kJmol(1)
Based on given Δr H
Δf H=HCO2=393.5kJmol1 . . . . (i)
ΔfH=H(H2O)=285.8kJmol1 . . . . (ii)
ΔfH=HO2=0.00(elements)
Required thermal reaction is for ΔfH of (CH4)
Thus, from III
890.3=[ΔfH(CH4)+2 Δf H(O2)]
[ΔfH(CO2)+2 ΔfH(H2O)]
=Δf H(CH4)+0[393.52×285.5]
ΔfH(CH4)=74.8 kJ/mol

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Thermochemistry
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon