Question

For the reaction; $H_2\left(g\right)+1/2O_2\left(g\right)=H_{2}O\left(l\right),△C_p=7.63cal/deg;△H_{{25}^{o}C}=68.3Kcal.$ what will be the value (in Kcal) of $△H$ at ${100}^{o}C$?

• A. $7.63\times (373-298)-68.3$
• B. $7.63\times {10}^{-3}\times (373-298)-68.3$
• C. $7.63\times {10}^{-3}\times (373-298)+68.3$
• D. $7.63\times (373-298)+68.3$

Answer 1

The correct option is B $7.63\times {10}^{-3}\times (373-298)-68.3$

Solution:-$(7.63\times {10}^{-3})\times (373-298)-68.3$

Given that:-

$\Delta H_{25℃}=-68.3\text{Kcal}$

$\Delta H_{100℃}=?$

$T_1=25℃=(273+25)=298K$

$T_2=100℃=(273+100)=373K$

$\Delta C_P=7.63cal/℃=7.63\times {10}^{-3}kcal/℃$

Using Kirchhoff's law,

$\Delta C_P=\frac{\Delta H_{100℃}-\Delta H_{25℃}}{T_2-T_1}$

$7.63\times {10}^{-3}=\frac{\Delta H_{100℃}-(-68.3\times {10}^3)}{373-298}$

$\Delta H_{100℃}+68.3=(7.63\times {10}^{-3})\times (373-298)$

$\Rightarrow \Delta H_{100℃}=(7.63\times {10}^{-3})\times (373-298)-68.3$

Hence the value of $\Delta H$ in $Kcal$ is $(7.63\times {10}^{-3})\times (373-298)-68.3=-67.72Kcal$.