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Calorimetry

At 298 K, t...

Question

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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Solution

The correct option is A 0
At 298 K, the enthalpy of combustion of $C H_{4}$ corresponds to the reaction
$C H_{4} (g) + 2 O_{2} (g) ⟶ C O_{2} (g) + 2 H_{2} O (l)$ .
Note the difference. Water is present in liquid state.
For the above reaction, $Δ_{c} H = - 890.3 k J / m o l$

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

Q. What volume of

H_{2} O_{(g)}

measured at STP is produced by the combustion of

8.00 g

of methane gas,

C H_{4}

, according to the following equation?

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

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

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}

H_{2} O

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

Q. The combustion of methane is written as

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

The difference between enthalpy change and energy change is equal to:

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