The enthalpy of combustion of methane,graphite and dihydrogen at $298 \, K$ are $-890.3 \, kJ \, mol^{-1}$,$-393.5 \, kJ \, mol^{-1}$ and $-285.8 \, kJ \, mol^{-1}$ respectively. The enthalpy of formation of $CH_{4(g)}$ will be:
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$
$(i) -74.8 \, kJ \, mol^{-1}$
$(ii) -52.27 \, kJ \, mol^{-1}$
$(iii) +74.8 \, kJ \, mol^{-1}$
$(iv) +52.26 \, kJ \, mol^{-1}$
(A) The combustion reactions are:
$(i) CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_{2}O_{(l)}; \Delta_cH = -890.3 \, kJ \, mol^{-1}$
$(ii) C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta_cH = -393.5 \, kJ \, mol^{-1}$
$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(l)}; \Delta_cH = -285.8 \, kJ \, mol^{-1}$
The formation reaction of $CH_{4(g)}$ is:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)}$
Using Hess's Law:
$\Delta_fH_{CH_4} = \Delta_cH_{C(s)} + 2(\Delta_cH_{H_2(g)}) - \Delta_cH_{CH_4(g)}$
$\Delta_fH_{CH_4} = [-393.5 + 2(-285.8) - (-890.3)] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = [-393.5 - 571.6 + 890.3] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = -74.8 \, kJ \, mol^{-1}$
Thus,the enthalpy of formation is $-74.8 \, kJ \, mol^{-1}$,which corresponds to option $(i)$.
$(i) CH_{4(g)} + 2O_{2(g)} \longrightarrow CO_{2(g)} + 2H_{2}O_{(l)}; \Delta_cH = -890.3 \, kJ \, mol^{-1}$
$(ii) C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}; \Delta_cH = -393.5 \, kJ \, mol^{-1}$
$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} \longrightarrow H_{2}O_{(l)}; \Delta_cH = -285.8 \, kJ \, mol^{-1}$
The formation reaction of $CH_{4(g)}$ is:
$C_{(s)} + 2H_{2(g)} \longrightarrow CH_{4(g)}$
Using Hess's Law:
$\Delta_fH_{CH_4} = \Delta_cH_{C(s)} + 2(\Delta_cH_{H_2(g)}) - \Delta_cH_{CH_4(g)}$
$\Delta_fH_{CH_4} = [-393.5 + 2(-285.8) - (-890.3)] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = [-393.5 - 571.6 + 890.3] \, kJ \, mol^{-1}$
$\Delta_fH_{CH_4} = -74.8 \, kJ \, mol^{-1}$
Thus,the enthalpy of formation is $-74.8 \, kJ \, mol^{-1}$,which corresponds to option $(i)$.
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Similar Questions
Calculate $\Delta H$ in $kJ$ for the following reaction:
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
$C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)}$
Given that:
$H_2O_{(g)} + C_{(s)} \longrightarrow CO_{(g)} + H_{2(g)} ; \Delta H = +131 \ kJ$
$CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta H = -282 \ kJ$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \longrightarrow H_2O_{(g)} ; \Delta H = -242 \ kJ$
DifficultAP EAMCET 2008
View SolutionFrom the following data at $25^{\circ} C$,calculate the $\Delta_{r} H^0$ for the reaction $H_2O_{(g)} \rightarrow 2 H_{(g)} + O_{(g)}$:
$1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ $\Delta H = 42.09 \ kJ \ mol^{-1}$ $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ $\Delta H = -242 \ kJ \ mol^{-1}$ $H_{2(g)} \rightarrow 2 H_{(g)}$ $\Delta H = 436 \ kJ \ mol^{-1}$ $O_{2(g)} \rightarrow 2 O_{(g)}$ $\Delta H = 496 \ kJ \ mol^{-1}$
| $1/2 H_{2(g)} + 1/2 O_{2(g)} \rightarrow OH_{(g)}$ | $\Delta H = 42.09 \ kJ \ mol^{-1}$ |
| $H_{2(g)} + 1/2 O_{2(g)} \rightarrow H_2O_{(g)}$ | $\Delta H = -242 \ kJ \ mol^{-1}$ |
| $H_{2(g)} \rightarrow 2 H_{(g)}$ | $\Delta H = 436 \ kJ \ mol^{-1}$ |
| $O_{2(g)} \rightarrow 2 O_{(g)}$ | $\Delta H = 496 \ kJ \ mol^{-1}$ |
The heat of combustion of acetaldehyde to carbon dioxide and water is $-1172 \ kJ \ mol^{-1}$. Calculate the amount of heat liberated when $66 \ g$ of acetaldehyde is completely oxidised. (Atomic mass: $C=12, H=1, O=16$) (in $kJ$)
DifficultMHT CET 2020
View SolutionWhich of the following has standard enthalpy of formation equal to zero?
Easy
View SolutionThe standard heat of formation,in $kcal / mol$ of $Ba^{2+}$ is
[Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$,
Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]
[Given : standard heat of formation of $SO_4^{2-}$ ion $(aq) = -216 \ kcal / mol$,
Standard heat of crystallisation of $BaSO_{4(s)} = -4.5 \ kcal / mol$,standard heat of formation of $BaSO_{4(s)} = -349 \ kcal / mol$]
MediumNEET 2025
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