Ncert Exercises & Solutions

Q.5. The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, –890.3 kJ $m o l^{- 1}$ –393.5 kJ $m o l^{- 1}$ , and –285.8 kJ $m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) –74.8 kJ $m o l^{- 1}$

(ii) –52.27 kJ $m o l^{- 1}$

(iii) +74.8 kJ $m o l^{- 1}$

(iv) +52.26 kJ $m o l^{- 1}$

NEETprep Answer:

According to the question,

(i) C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}

∆ H = - 890.3 k J m o l^{- 1}

(i i) C_{(s)} + O_{2 (g)} \to C O_{2 (g)}

∆ H = - 393.5 k J m o l^{- 1}

(i i i) 2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(g)}

∆ H = - 285.8 k J m o l^{- 1}

Thus, the desired equation is the one that represents the formation of

C H_{4 (g)}

i.e.,

C_{(s)} + 2 H_{2 (g)} \to C H_{4 (g)} ∆_{f} H_{C H_{4}} = ∆_{c} H_{c} + 2 ∆_{c} H_{H_{2}} - ∆_{c} H_{C O_{2}} = [- 393.5 + 2 (- 285.8) - (- 890.3)] k J m o l^{- 1} = - 74.8 k J m o l^{- 1}

∴

Enthalpy of formation of

C H_{4 (g)} = - 74.8 k J m o l^{- 1}

Hence, alternative (i) is correct.