The enthalpy change for the following reaction, is called:
$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 {C O}_{2} (g) + 2 H_{2} O (g)$

enthalpy of formation

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enthalpy of combustion

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enthalpy of vaporisation

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enthalpy of sublimation

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Solution

The correct option is B enthalpy of combustion
Enthalpy of combustion is defined as the energy change of any reaction involving 1 mole of species reacting with sufficient oxygen.
Since, in above reaction 1 mole of $C_{2} H_{4}$ is reacting with sufficient $O_{2}$ to form $C O_{2} and H_{2} O .$
Hence, answer is option $B$ .

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Similar questions

Q. The enthalpy change for the reaction,

C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (g)

is called:

Q. The enthalpy change for the reaction,

C_{2} H_{4} (g) + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (g)

is called:

Q. The enthalpy change for the reaction,

(g) C_{2} H_{4} + 3 O_{2} (g) ⟶ 2 C O_{2} (g) + 2 H_{2} O (g)

is called:

For the reaction $C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)$ , the difference between enthalpy change and internal energy change is:

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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The enthalpy change for the following reaction, is called:C2H4(g)+3O2(g)→2CO2(g)+2H2O(g)

The correct option is B enthalpy of combustionEnthalpy of combustion is defined as the energy change of any reaction involving 1 mole of species reacting with sufficient oxygen.Since, in above reaction 1 mole of C2H4 is reacting with sufficient O2 to form CO2 and H2O.Hence, answer is option B.

The enthalpy change for the following reaction, is called:
$C_{2} H_{4} (g) + 3 O_{2} (g) \to 2 {C O}_{2} (g) + 2 H_{2} O (g)$