Calculate the mass of mercury which can be liberated from $H g O$ at $25^{\circ} C$ by the treatment of excess $H g O$ with $41.84 k J$ of heat at
(i) constant pressure (ii) constant volume conditions.
Given: $△ H_{f}^{\circ} (H g O, s) = - 90 k J m o l^{- 1}$ and $M (H g) = 200.6 g m o l^{- 1}$ .

Open in App

Solution

${\frac{1}{2} O_{2}}_{(g)} + {H g}_{(l)} ⟶ {H g O}_{(s)} Δ H = - 90$ $k J / m o l$

$(i)$ $1$ $m o l$ of $H g ⟶ 90$ $k J$

$∴ x$ $m o l$ of $H g ⟶ 41.84$ $k J$

$∴ x = \frac{41.84}{90} = 0.465$ $m o l$

$∴$ Mass of mercury $= 0.465 \times 200.6 = 93.28$ $g$

$(i i)$ At constant volume, $q = Δ V = Δ H - Δ n_{g} R T$

$∴ q = - 90 - Δ n_{g} R T$

Here, $Δ n_{g} =$ Difference between gaseous moles of reactants and products

$⟹ Δ n_{g} = 0 - \frac{1}{2} = - \frac{1}{2}$

$∴ q = - 90 + \frac{1}{2} \times 8.314 \times 298 \times 10^{- 3} = - 88.76$ $k J$

$∴ M o l e s$ of $H g = \frac{41.84}{88.76} = 0.471$ $m o l$

$∴ M a s s$ of $H g = 0.471 \times 200.6 \approx 94.5$ $g$

Suggest Corrections

Similar questions

H g O (s)

- 90.8 k J m o l^{- 1}

H g O (s)

H g = 200

H g O

H g O

H_{2} O + H g O + 4 I^{-} \to H g I_{4}^{2 -} + 2 O H^{-}

H g O

H g O + 4 I^{-} ⟶ H g I_{4}^{2 -} + 2 {O H}^{-}

H g O \equiv 2 {O H}^{-} \equiv 2 H^{+}

x

H g O

10

x

Given that,

$Z n + \frac{1}{2} O_{2} \to Z n O + 35.25 K J$

$H g O \to H g + \frac{1}{2} O_{2} - 9.11 K J$

The heat of the reaction $Z n + H g O \to Z n O + H g$ is

Z n + 1 / 2 O_{2} \to Z n O + 35.25 k J . H g O \to H g + 1 / 2 O_{2} + 9.11 k J .

Z n + H g O \to Z n O + H g

Join BYJU'S Learning Program