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  • At  and 1 atm pressure, the enthalpies of combustion are as give

At \mathrm{25^{\circ}C} and 1 atm pressure, the enthalpies of combustion are as given below:

Substance \mathrm{H_{2}} C (graphite) \mathrm{C_{2}H_{6}(g)}
\mathrm{\frac{\Delta _{c}H^{\Theta }}{kJ\;mol^{-1}}} -286.0 -394.0 -1560.0

The enthalpy of formation of ethane is

Option: 1

\mathrm{+54.0\; kJ\;mol^{-1}}


Option: 2

\mathrm{-68.0\; kJ\;mol^{-1}}


Option: 3

\mathrm{-86.0\; kJ\;mol^{-1}}


Option: 4

\mathrm{+97.0\; kJ\;mol^{-1}}


Answers (1)

best_answer

Given,

\begin{aligned} &\mathrm{H}_{2}+\mathrm{\frac{1}{2}} \mathrm{O}_{2} \rightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{\Delta }_{\mathrm{c}} \mathrm{H}=-286 \mathrm{~kJ/} \mathrm{\textrm {mol }}----(1) \\ &\mathrm{C}_{(\mathrm{s})}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} \quad \mathrm{\Delta }_{\mathrm{c}} \mathrm{H}=-394 \mathrm{~kJ} / \mathrm{mol} -----(2)\\ & \end{aligned}

\begin{aligned} &\mathrm{C}_{2} \mathrm{H}_{6}+\frac{7}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O} \quad \Delta_{\mathrm{C}} \mathrm{H}=-1560 \mathrm{~kJ}\: / \mathrm{mol} ----- \text { (3) } \end{aligned}

The enthapy of formation of ethane :

\mathrm{2 C_{(s)}+3 H_{2} \rightarrow C_{2} H_{6} \quad \Delta _{C} H=\text { ? }}

The formation reaction of ethane can be obtained by 2 \times(2)+3 \times(1)-(3)

Then ,

\begin{aligned} \Delta_{f} \mathrm{H} \text { of } \mathrm{C}_{2} \mathrm{H}_{6}=& 2 \times(-394)+3 \times \mathrm{(-286)}-(-1560) \\ =&-1646+1560\\=&-86 \mathrm{~kJ} / \mathrm{mol} \end{aligned}

Hence, the correct answer is Option (3)

Posted by

Pankaj

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