
If \[{x_1}\] , \[{x_2}\] and \[\;{x_3}\] are enthalpies of \[H-H\], \[O = O\] and \[O-H\] bonds respectively, and \[{x_4}\] is the enthalpy of vaporization of water, estimate the standard enthalpy of combustion of hydrogen.
(a) \[{x_1} + {x_2}/2 - 2{x_3} + {x_4}\]
(b) \[{x_1} + {x_2}/2 - 2{x_3} - {x_4}\]
(c) \[{x_1} + {x_2}/2 - {x_3} + {x_4}\]
(d) \[2{x_3} - {x_1} - {x_2}/2 - {x_4}\]
Answer
532.2k+ views
Hint: Combustion of hydrogen needs to be checked and all the combustion enthalpies related to that need to be noted for further steps. Calculations should be done using the formula of combustion enthalpy of hydrogen.
Complete step-by-step answer: Combustion of hydrogen:
$ \Rightarrow $ \[{H_2} + \dfrac{1}{2}{O_2} \to {H_2}O\;\] ( \[H - O - H\] )
As we know, water has $2$ $O - H$ bonds.
Therefore, combustion enthalpy of hydrogen:
\[\;\; \Rightarrow \;\Delta H\]\[ = \]Bond energy of reactant − Bond energy of the product − Enthalpy of evaporation.
Now, we can write it as,
$ \Rightarrow $$\Delta $
Additional information:
Enthalpy \[H\] is stated as the sum of the internal energy \[U\] and the product of pressure and volume \[\;PV\] known by the equation: \[H = U + PV\]. Whenever a process takes place at the same pressure, the heat evolved (either released or absorbed) is the same as the change in enthalpy.
The enthalpy of combustion of a substance is stated as the heat energy which is given out when \[1\] mole of a substance burns completely in oxygen gas.
Enthalpy can be calculated by working out the no. of moles of fuel which has been burnt. This is generally done by working out the difference b/w the mass of fuel before the water gets heated, and the mass of fuel afterwards. Then work out the energy given out by one mole of the substance. This will provide the \[J/mo{l^{ - 1}}\] , therefore to get the answer in \[KJ\], divide it by \[1000\]
Hence, Option B is correct.
Note: Carefully observe the hint and from there it is quite clear to find the enthalpy first using the formula of enthalpy of hydrogen and then by checking the number of moles exactly. Later relate it with the options to find the correct answer.
Complete step-by-step answer: Combustion of hydrogen:
$ \Rightarrow $ \[{H_2} + \dfrac{1}{2}{O_2} \to {H_2}O\;\] ( \[H - O - H\] )
As we know, water has $2$ $O - H$ bonds.
Therefore, combustion enthalpy of hydrogen:
\[\;\; \Rightarrow \;\Delta H\]\[ = \]Bond energy of reactant − Bond energy of the product − Enthalpy of evaporation.
Now, we can write it as,
$ \Rightarrow $$\Delta $
Additional information:
Enthalpy \[H\] is stated as the sum of the internal energy \[U\] and the product of pressure and volume \[\;PV\] known by the equation: \[H = U + PV\]. Whenever a process takes place at the same pressure, the heat evolved (either released or absorbed) is the same as the change in enthalpy.
The enthalpy of combustion of a substance is stated as the heat energy which is given out when \[1\] mole of a substance burns completely in oxygen gas.
Enthalpy can be calculated by working out the no. of moles of fuel which has been burnt. This is generally done by working out the difference b/w the mass of fuel before the water gets heated, and the mass of fuel afterwards. Then work out the energy given out by one mole of the substance. This will provide the \[J/mo{l^{ - 1}}\] , therefore to get the answer in \[KJ\], divide it by \[1000\]
Hence, Option B is correct.
Note: Carefully observe the hint and from there it is quite clear to find the enthalpy first using the formula of enthalpy of hydrogen and then by checking the number of moles exactly. Later relate it with the options to find the correct answer.
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