Question

calculate the heat energy released when 26.0 g of liquid mercury at 25.00 °c is converted to solid mercury at its melting point. constants for mercury at 1 atm heat capacity of hg(l) 28.0 j/(mol k) melting point 234.32 k enthalpy of fusion 2.29 kj/mol

Explanation:

Step1: Convert mass of mercury to moles

The molar mass of mercury (Hg) is $M = 200.59\ g/mol$. The number of moles $n$ of mercury is calculated by $n=\frac{m}{M}$, where $m = 26.0\ g$. So $n=\frac{26.0\ g}{200.59\ g/mol}= 0.1296\ mol$.

Step2: Calculate heat released during cooling

First, convert the initial temperature $T_1=25.00^{\circ}C=(25.00 + 273.15)K=298.15\ K$. The melting - point $T_2 = 234.32\ K$. The heat released during cooling $q_1$ is given by $q_1=n\times C\times\Delta T$, where $C = 28.0\ J/(mol\cdot K)$ and $\Delta T=T_1 - T_2$. So $q_1=0.1296\ mol\times28.0\ J/(mol\cdot K)\times(298.15\ K - 234.32\ K)=0.1296\ mol\times28.0\ J/(mol\cdot K)\times63.83\ K = 232.7\ J$.

Step3: Calculate heat released during freezing

The heat released during freezing $q_2$ is given by $q_2=n\times\Delta H_{fus}$, where $\Delta H_{fus}=2.29\ kJ/mol$. So $q_2=0.1296\ mol\times2.29\ kJ/mol = 0.2968\ kJ=296.8\ J$.

Step4: Calculate total heat released

The total heat released $q = q_1+q_2$. First, convert $q_1$ to $kJ$: $q_1 = 232.7\ J=0.2327\ kJ$. Then $q=0.2327\ kJ + 0.2968\ kJ=0.5295\ kJ$.

Answer:

$0.530$