$H _{2}+\frac{1}{2} O _{2} \rightarrow H _{2} O , \cdots \cdots( ii )$ $\Delta H =-\,287.3 \,kJ\,mol ^{-1}$
$2 CO _{2}+3 H _{2} O \rightarrow C _{2} H _{5} OH +3 O _{2} \cdots \cdots ( iii )$; $ \Delta H =1366.8 \,kJ\,mol ^{-1}$
$C _{2} H _{5} OH (1)$ માટે ની રચનાની પ્રમાણિત એન્થાલ્પી શોધો
The enthalpy change for the above reaction is \(-\,393\, kJ \,mol ^{-1}\)
\(H _{2}+\frac{1}{2} O _{2} \rightarrow H _{2} O \cdots \cdots( II )\)
The enthalpy change for the above reaction is \(-287.3\, kJ \,mol ^{-1}\)
\(2 CO _{2}+3 H _{2} O \rightarrow C _{2} H _{5} OH +3 O _{2} \cdots \cdots( III )\)
The enthalpy change for the above reaction is \(1366.8\,kJ \,mol ^{-1}\)
The formation reaction of \(C _{2} H _{5} OH (l)\) is as follows :
\(2 C +3 H _{2}+\frac{1}{2} O _{2} \rightarrow C _{2} H _{5} OH\)
To derive the above reaction, multiply reaction \((I)\) by \(2\), multiply reaction \((II)\) by \(3\) and then add all the reactions.
Therefore, the total change in enthalpy will become as :
\(\Delta H=-\,393\, kJ \,mol ^{-1} \times 2+\left(-287.3 \,kJ \,mol ^{-1}\right) \times 3+\)\(1366.8 \,kJ \,mol ^{-1}\)
\(\Delta H=-\,281.1 \,kJ \,mol ^{-1}\)
$(i)\,\,\Delta H_f^o\,\,of\,{H_2}{O_{(\ell )}}\, = \,\, - 68.3\,K\,\,cal\,\,mo{l^{ - 1}}$
$(ii)\,\,\Delta H_{comb}^o\,\,of\,{C_2}{H_2}\, = \,\, - 337.2\,K\,\,cal\,\,mo{l^{ - 1}}$
$(iii)\,\,\Delta H_{comb}^o\,\,of\,\,{C_2}{H_4}\,\, = \,\, - \,363.7\,\,K\,\,cal\,\,mo{l^{ - 1}} $
$(A)$ $2 CO ( g )+ O _2( g ) \rightarrow 2 CO _2( g ) \quad \Delta H _1^\theta=- x\,kJ\,mol { }^{-1}$
$(B)$ $C$ (graphite) $+ O _2$ (g) $\rightarrow CO _2$ (g) $\Delta H _2^\theta=- y\,kJ\,mol -1$
$C$(ગ્રેફાઈટ) $+$ $\frac{1}{2} O _2( g ) \rightarrow CO ( g )$ પ્રક્રિયા માટે $\Delta H ^\theta$ શોધો.