Question

When solid ${NH}_4{NO}_3$ is dissolved in water at ${25}^{o}C$, the temperature of the solution decreases. What is true about the signs of $\Delta H$ and $\Delta S$ for this process__________.

• A. $\Delta H=-ve,\Delta S=+ve$
• B. $\Delta H=-ve,\Delta S=-ve$
• C. $\Delta H=+ve,\Delta S=+ve$
• D. $\Delta H=+ve,\Delta S=-ve$

Answer 1

The correct option is C $\Delta H=+ve,\Delta S=+ve$

$\text{1.Since temperature of the solution decreases means it's an exothermic reaction so}$ $\left(\Delta H\right)=+ve$

$\text{2.Since solid}$ $N{H}_{4}N{O}_3$ $\text{is dissociated so number of ions increases so randomness increases and hence}$ $\left(\Delta S\right)=+ve$