Calculate the enthalpy change for the following reaction
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
Given enthalpies of formation of $C H_{4}, C O_{2} & H_{2} O$ are $- 74.8 K J m o l^{- 1}, - 393.5 K J m o l^{- 1}$ and $- 286.2 K J m o l^{- 1}$ respectively.

891.1 KJ

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-891.1 KJ

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890.1 KJ

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-890.1 KJ

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Solution

The correct option is B
-891.1 KJ

The change in enthalpy can be given by:
$△ H^{o} = △ H_{f (p r o d u c t s)}^{o} - △ H_{f (r e a c t a n t s)}^{o}$
$= [△ H_{f (C O_{2})}^{o} + 2 △ H_{f (H_{2} O)}^{o}] - [△ H_{f (C H_{4})}^{o} + 2 △ H_{f (O_{2})}^{o}]$
$= [- 393.5 + 2 \times (- 286.2)] - [- 74.8 + 2 \times 0]$

$= [- 393.5 + 2 \times (- 286.2)] - [- 74.8 + 2 \times 0]$
$△ H o = - 891.1 K J$

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Similar questions

Q. Calculate the enthalpy change for the following reaction:

{C H}_{4} (g) + 2 O_{2} (g) ⟶ {C O}_{2} (g) + 2 H_{2} O (l)

Given, enthalpies of formation of

{C H}_{4}, {C O}_{2}

and

H_{2} O

are

74.8 k J {m o l}^{- 1}, - 393.5 k J {m o l}^{- 1}, - 286 k J {m o l}^{- 1}

respectively.

Standard enthalpy of combustion of $C H_{4}$ is $- 890 k J m o l^{- 1}$ and standard enthalpy of vaporisation of water is 40.5 kJ $m o l^{- 1}$ . Calculate the enthalpy change for the reaction :
$C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(g)}$

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. Enthalpy of formation of CH₄₍_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.

Q. At

298 K

, the enthalpy of combustion of

C H_{4}

corresponds to the reaction

C H_{4} (g) + 2 O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (g)

.If true enter 1 else o

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Calculate the enthalpy change for the following reaction CH4(g) + 2O2(g)→CO2(g) + 2H2O(l) Given enthalpies of formation of CH4, CO2 & H2O are −74.8KJ mol−1,−393.5KJ mol−1 and −286.2KJ mol−1 respectively.

The correct option is B -891.1 KJ The change in enthalpy can be given by: △Ho=△Hof(products)−△Hof(reactants) =[△Hof(CO2)+2△Hof(H2O)]−[△Hof(CH4)+2△Hof(O2)] =[−393.5+2×(−286.2)]−[−74.8+2×0] =[−393.5+2×(−286.2)]−[−74.8+2×0] △Ho=−891.1KJ

Calculate the enthalpy change for the following reaction
$C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$
Given enthalpies of formation of $C H_{4}, C O_{2} & H_{2} O$ are $- 74.8 K J m o l^{- 1}, - 393.5 K J m o l^{- 1}$ and $- 286.2 K J m o l^{- 1}$ respectively.