The enthalpy of vapourisation of liquid water using the data H2(g)+12O2(g)→H2O(l);ΔH=−285.77KJmol−1 H2(g)+12O2(g)→H2O(g);ΔH=−241.84KJmol−1
A
+43.93 KJ mol−1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
-43.93 KJ mol−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
+527.61 KJ mol−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
-527.61 KJ mol−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A +43.93 KJ mol−1 The thermochemical reactions are as given below. H2(g)+12O2(g)→H2O(l);ΔH=−285.77KJmol−1 ......(1) H2(g)+12O2(g)→H2O(g);ΔH=−241.84KJmol−1 ......(2) The reaction (1) is reversed and added to the reaction (2) to obtain the reaction for the vapouisation of liquid water which is H2O(l)→H2O(g) Hence, the heat of vapourisation of liquid water is −241.84+285.77=43.93KJmol−1