Question

∆H of water is -285.8 kJ mol-1, If enthalpy of neutralization of monoacid strong base is -57.3 kJ per mole ∆h of
of OH-ion will be
(1) -228.5 kJ mol-1
(2) 228.5 kJ mol-1
(3) 114.25 kJ mol-1
(4) -114.25 kJ mol-1​

Answer 1

Answer:

-114.25 KJmol-1is the answer

Answer 2

Answer:

a

Explanation:

H2+12O2→H2O ΔH=−285.8  

Target reaction H++OH−→H2O ΔH=−57.3

Δ=−57.3=[−285.8−0−x] x=ΔH∘f of OH−

x=−228.5 kJ mol−1