H2+1/2O2⟶H2O;ΔH⊖=−68Kcal. K+H2O+aq⟶KOH(aq)+1/2H2;ΔH⊖=−48Kcal. KOH+aq⟶KOH(aq);ΔH⊖=−14Kcal From the above data, the standard heat of formation of KOHinKcal is
A
−68+48−14
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B
−68−48+14
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C
68−48+14
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D
68+48+14
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Solution
The correct option is C−68−48+14 H2+1/2O2⟶H2O ...(i) K+H2O+aq⟶KOH(aq)+1/2H2 ...(ii) KOH+aq⟶KOH(aq) ...(iii) Equation (ii)+ Equation (i)- Equation (iii) gives K(s)+12O2(g)+12H2(g)→KOH(g) ΔH=−48+(−68)−(−14)=−68−48+14