2 mol of Hg_{(g)} is combusted in a fixed vo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The heat released in the bomb calorimeter is $Q = C \Delta T = 20.00 \ kJ \ K^{-1} 	imes (312.8 - 298.0) \ K = 20.00 	imes 14.8 = 296 \ kJ$.Sinc

Vedclass · Accepted Answer

$90.39$

Vedclass · Answer

(A) The heat released in the bomb calorimeter is $Q = C \Delta T = 20.00 \ kJ \ K^{-1} 	imes (312.8 - 298.0) \ K = 20.00 	imes 14.8 = 296 \ kJ$.Since $2 \ mol$ of $Hg_{(g)}$ are reacted,the internal energy change for the reaction $Hg_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow HgO_{(s)}$ is $\Delta U = -\frac{296 \ kJ}{2 \ mol} = -148 \ kJ \ mol^{-1}$.Using $\Delta H = \Delta U + \Delta n_g RT$,where $\Delta n_g = 0 - (1 + 0.5) = -1.5 \ mol$:$\Delta H = -148 + (-1.5 	imes 8.3 	imes 10^{-3} 	imes 298) = -148 - 3.7101 = -151.7101 \ kJ \ mol^{-1}$.This is the enthalpy change for the reaction $Hg_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow HgO_{(s)}$.Given $\Delta H_f(Hg_{(g)}) = 61.32 \ kJ \ mol^{-1}$,we use the relation $\Delta H_{reaction} = \Delta H_f(HgO_{(s)}) - [\Delta H_f(Hg_{(g)}) + \frac{1}{2} \Delta H_f(O_{2(g)})]$.$-151.7101 = \Delta H_f(HgO_{(s)}) - 61.32$.$\Delta H_f(HgO_{(s)}) = -151.7101 + 61.32 = -90.39 \ kJ \ mol^{-1}$.Thus,$|X| = 90.39$.

Similar Questions

The work done during the combustion of $9 \times 10^{-2} \ kg$ of ethane,$C_2H_{6(g)}$ at $300 \ K$ is (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,atomic mass $C = 12$,$H = 1$). (in $kJ$)

The enthalpy of combustion of carbon monoxide at $17 ^\circ C$ and constant volume is $-283.3 \ kJ$. Calculate the enthalpy of combustion at constant pressure in $kJ$.

At $300 \, K$ temperature,the standard heat of formation of benzoic acid $(s)$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-408$,$-393$,and $-286 \, kJ \, mol^{-1}$ respectively. At constant pressure,the heat of combustion for benzoic acid will be $..... \, kJ$.

For the reaction $\frac{1}{2}X_2 + \frac{3}{2}Y_2 \to XY_3$,$\Delta H = -30 \ kJ/mol$. Given $\Delta S_{X_2} = 60 \ J/mol \cdot K$,$\Delta S_{Y_2} = 40 \ J/mol \cdot K$,and $\Delta S_{XY_3} = 50 \ J/mol \cdot K$,calculate the temperature at equilibrium in $K$.

The ${\Delta _c}{U^\circ}$ of combustion of $CH_{4(g)}$ at a certain temperature is $-100 \ kJ/mol$. The value of ${\Delta _c}{H^\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module