Question

Calculate the entropy change in surroundings when 1.00 mol of $H_{2}O\left(l\right)$ is formed under standard conditions at 298 K. Given ${\Delta }_r{H}^0$ = - 286 kJ $mo{l}^{-1}$.

Answer 1

${H_2}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{H_{2}O}_{\left(l\right)}{\Delta }_f{H}^0=-286KJ/mol$

From the above equation,

At $298K$, when $1$ mole of ${H_{2}O}_{\left(l\right)}$ is formed, $286KJ$ of heat is released. The same amount of heat is absorbed by the surroundings.

$\therefore q_{surr.}=+286KJ/mol;T=298K$

As we know that,

${\Delta S}_{surr.}=\frac{q_{surr.}}{T}$

$\therefore {\Delta S}_{surr.}=\frac{286}{298}=0.96KJ/mol-K$

Hence the entropy change in surroundings will be $0.96KJ/mol$.