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.

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Solution

$Δ H^{o} = Δ H_{f (p r o d u c t s)}^{o} - Δ H_{f (r e a c t a n t s)}^{o}$

$= [Δ H_{({C O}_{2})}^{o} + Δ H_{(H_{2} O)}^{o}] - [Δ H_{({C H}_{4})}^{o} + 2 Δ H_{(O_{2})}^{o}]$

$= [- 393.5 + 2 \times (- 286.2)] - [- 74.8 + 2 \times 0] = - 890.7 k J$

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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.

Calculate the standard enthalpy of formation of CH₃OH₍_l₎ from the following data:

CH₃OH₍_l₎ + O₂₍_g₎ CO₂₍_g₎ + 2H₂O₍_l₎ ; Δ_rH^θ = –726 kJ mol^–1

C₍_g₎ + O₂₍_g₎ CO₂₍_g₎ ; Δ_cH^θ = –393 kJ mol^–1

H₂₍_g₎ +O₂₍_g₎ H₂O₍_l₎ ; Δ_fH^θ = –286 kJ mol^–1.

Q. Calculate the standard enthalpy of formation of

C H_{3} O H (I)

from the following data :

C H_{3} O H (I) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l); △_{r} H^{θ} = - 726 k J m p l^{- 1}

C_{(g r a p h i t e)} + O_{2} (g) \to C O_{2} (g); △_{c} H^{θ} = - 393 k J m o l^{- 1}

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △_{f} H^{θ} = - 286 k J m o l^{- 1}

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.

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

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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 and H2O are 74.8 kJmol−1,−393.5 kJmol−1,−286 kJmol−1 respectively.

ΔHo=ΔHof(products)−ΔHof(reactants)=[ΔHo(CO2)+ΔHo(H2O)]−[ΔHo(CH4)+2ΔHo(O2)]=[−393.5+2×(−286.2)]−[−74.8+2×0]=−890.7kJ

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.

$Δ H^{o} = Δ H_{f (p r o d u c t s)}^{o} - Δ H_{f (r e a c t a n t s)}^{o}$

$= [Δ H_{({C O}_{2})}^{o} + Δ H_{(H_{2} O)}^{o}] - [Δ H_{({C H}_{4})}^{o} + 2 Δ H_{(O_{2})}^{o}]$

$= [- 393.5 + 2 \times (- 286.2)] - [- 74.8 + 2 \times 0] = - 890.7 k J$