DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -352.8 kJ. Delta H_(f)^(0) for HF is -268.3 KJ mol^(-1), then Delta H_(f)^(0) of HCl would be

DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -

1.	DeltaH^(0) for a reaction F_(2) + 2HCl rarr 2HF + Cl_(2) is given as -352.8 kJ. Delta H_(f)^(0) for HF is -268.3 KJ mol^(-1), then Delta H_(f)^(0) of HCl would be
Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`Delta <a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a> = H _(P) - H _(R) = <a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a> H _(<a href="https://interviewquestions.tuteehub.com/tag/hf-479831" style="font-weight:bold;" target="_blank" title="Click to know more about HF">HF</a>) - <a href="https://interviewquestions.tuteehub.com/tag/2h-300377" style="font-weight:bold;" target="_blank" title="Click to know more about 2H">2H</a> _(HCl)` <br/> `=- 352.8 kJ = 2 (0268.3) - 2 (H _(HCl))` <br/> `2 (H _(HCl))=-536.6 + 352.8 =-183.8kJ.` <br/> (or) `H _(HCl) =-91.9 kJ mol ^(-1).`</body></html>