Question

At ${25}^{0}C$ the enthalpy change for the reaction $H_{2}S{O}_4+5H_{2}O=H_{2}S{O}_4.5H_{2}O$ (all liquids) is -580.32 kj / mol. Calculate the temperature change if 1 mole of $H_{2}S{O}_4$ is dropped in to 5 mole of $H_{2}O$ at ${25}^{0}C$. Assume no heat loss to the surroundings and that the specific heat capacity of solution is 4.184 j $k^{-1}{g}^{-1}$.

Answer 1

$1$ $mole$ of $H_{2}S{O}_4\cdot 5H_{2}O$ will be formed.

Heat absorbed $=Ne\cdot \Delta H$ for $1$ $mole=580.32$ $kJ$

$1$ $mole=188$ $g$ of $H_{2}S{O}_4\cdot 5H_{2}O$

$\therefore 580.32\times {10}^3=188\times 4.184\times \Delta T$

$\therefore \Delta T=737$ $K$