The enthalpy of formation of methane at constant pressure and 300 K is – 78.84 KJ. The enthalpy of formation at constant volume is:

– 75.5 kJ

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– 102.6 kJ

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+ 75.5 kJ

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– 54.0 kJ

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Solution

The correct option is A – 75.5 kJ
Basic ideas to keep in mind while solving this numerical are:
$q_{p} = Δ H, q_{v} = Δ U$

The equation representing the enthalpy of formation of methane is :
$C (s) + 2 H_{2} (g) \to C H_{4} (g); Δ H = - 78 k J Δ H = 78 k J; Δ n g = 1 - 2 = - 1 m o l R = 8.314 \times 10^{- 3} k J K^{- 1}; T = 300 K A c c o r d i n g t o t h e r e l a t i o n, Δ H = Δ U + Δ n g R T Δ U = Δ H - Δ n g R T = (- 78 k J) - (- 1 m o l) \times (8.314 \times 10^{3} K J k^{- 1} m o l^{- 1}) \times 300 K = - 78.0 + 2.49 = - 75.5 k J$

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Similar questions

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, $- 890.3 k J m o l^{- 1} - 393.5 k J m o l^{- 1}$ , and $- 285.8 k J m o l^{- 1}$ respectively. Enthalpy of formation of $C H_{4 (g)}$ will be
(i) $- 74.8 k J m o l^{- 1}$
(ii) $- 52.27 k J m o l^{- 1}$
(iii) $+ 74.8 k J m o l^{- 1}$
(iv) $+ 52.26 k J m o l^{- 1}$

Q. The enthalpy of combustion of methane graphite and dihydrogen at

298 K

are

- 890.3

m o l^{- 1}

and

- 285.8

m o l^{- 1}

respectively. Enthalpy of formation of

C H_{4} (g)

will be:

Q. The enthalpy of formation of methane at constant pressure and 300 K is – 78.84 KJ. The enthalpy of formation at constant volume is:

The enthalpy of formation of methane at constant pressure and $300 K$ is $- 75.83 K J$ . What will be the heat of formation at constant volume? $(R = 8.3 J K^{- 1} m o l^{- 1}$ )

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