Question

Calculate the enthalpy change for the process,
$H_{2}O(l,-{10}^{o}C)$ $⟶$ $H_{2}O(s,-{10}^{o}C)$
given that
$C_p(H_{2}O,l)=$ 75.4 $J{K}^{-1}mo{l}^{-1}$
$C_p(H_{2}O,s)=$ 37.2 $J{K}^{-1}mo{l}^{-1}$
$H_{2}O(l,0^{o}C)$ $⟶$ $H_{2}O(s,0^{o}C)$, $\Delta H=$ -6008 $Jmo{l}^{-1}$

• A. -5626 $Jmo{l}^{-1}$
• B. 5626 $Jmo{l}^{-1}$
• C. 5.626 $Jmo{l}^{-1}$
• D. 56.26 $Jmo{l}^{-1}$

Answer 1

The correct option is A -5626 $Jmo{l}^{-1}$

According to Kirchhoff's Law:
$\Delta H_2-\Delta H_1=$ $\left[C_p\right(products)-C_p(reactants\left)\right]$ $(T_2-T_1)$
where, $\Delta H_2=$ enthaply of reaction at temperature $T_2$
$\Delta H_1=$ enthaply of reaction at temperature $T_1$
$C_p$ = heat capacity at constant pressure
$T_2=$ $-{10}^{o}C$
so putting values of all these in above equation we get:
$\Delta H_2-(-6008)=(37.2-75.4)\times (-10-0)$
$\Delta H_2+6008=$ $-38.2(-10)$
$\Delta H_2=$ $-5626$ $Jmo{l}^{-1}$