\(q = 16\)
\({C_P} = 24\)
\({C_P} = \frac{{qP}}{{\Delta T}}\)
\(\Delta T = \frac{{16}}{{24}}\,K = \frac{2}{3}\,K\)
(Given ${\Delta _{fus}}H = 6\, kJ\, mol^{-1}$ at $0\,^oC$,
$C_p(H_2O, l) =75.3\, J\, mol^{-1} \, K^{-1}$ ,
$C_p(H_2O, s) = 36.8\, J\, mol^{-1} \, K^{ -1}$ )
$A.$ $I _2( g ) \rightarrow 2 I ( g )$
$B.$ $HCl ( g ) \rightarrow H ( g )+ Cl ( g )$
$C.$ $H _2 O ( l ) \rightarrow H _2 O ( g )$
$D.$ $C ( s )+ O _2( g ) \rightarrow CO _2( g )$
$E.$ પાણીમાં એમોનિયમ કલોરાઈડનું વિલયન (ઓગળવું)
$\Delta H = - 98.7\,{\mkern 1mu} kJ{\mkern 1mu} {\mkern 1mu} S{O_3} + {H_2}O \to {H_2}S{O_4};\Delta H = - 130.2{\mkern 1mu} \,kJ;$
${H_2} + \frac{1}{2}{\mkern 1mu} {O_2} \to {H_2}O;{\Delta _H} = - 287.3{\mkern 1mu} \,kJ$
તો $298\, K$ એ $H_2SO_4$ ની નિર્માણ એન્થાલ્પી ............. $\mathrm{kJ}$ માં શોધો.