Question

At 300K the heat of combustion of methane in a bomb calorimeter is $ {{ - 900}} $ KJ/mol, so its enthalpy of combustion in an open flask will be approximately equal to what?
(A) $ {{ - 850}} $ KJ/mol
(B) $ {{ - 950}} $ KJ/mol
(C) $ {{ - 900}} $ KJ/mol
(D) Zero

Answer 1

Hint: In the above question, we have to find the enthalpy of combustion when heat of combustion ( $ {{\Delta U}} $ ) is given. We have to first write a balanced chemical equation of combustion of methane and then we can use the relation between enthalpy change and internal energy change to solve this question.

Formula Used
 $ {{\Delta H = \Delta U + W}} $
Where $ {{\Delta H}} $ = enthalpy change
 $ {{\Delta U}} $ =internal energy change
W= work done.

Complete step by step solution
We know that work done is equal to product of pressure and volume change and hence, we can write W= P $ {{\Delta }} $ V
But in the question, pressure and volume is not given. So, we can use the ideal gas equation for this.
According to ideal gas equation:
 $ {{PV = nRT}} $
Which implies that:
 $ {{P}}\Delta {{V = }}\Delta {{nRT}} $
Where $ {{\Delta n}} $ = change in number of moles of gaseous substance.
R= universal gas constant.
T= temperature in K
We can now rewrite the enthalpy equation as:
 $ {{\Delta H = \Delta U + }}\Delta {{nRT}} $ ............................(1)
Let us now write a balanced chemical equation of combustion of methane:
 $ {{C}}{{{H}}_{{4}}}{{(g) + 2}}{{{O}}_{{2}}}{{(g)}} \to {{C}}{{{O}}_{{2}}}{{(g) + 2}}{{{H}}_{{2}}}{{O(l)}} $
 $ {{\Delta n}} $ = number of moles of gaseous product – number of moles of gaseous reactant
 $ {{\Delta n = }}1 - (2 + 1) = - 2 $
Substituting the values in equation 1, we get:
 $ {{\Delta H = \Delta U + }}\Delta {{nRT}} $
 $ {{\Delta H}} = - 900 + ( - 2) \times 0.0083 \times 300 = - 900 - 5 = - 905 $ KJ/mol which is nearly equal to $ {{ - 900}} $ kJ/mol.
Hence, the correct option is option C.

Note
In these type of questions where heat of formation, combustion and etc. are given, and we are asked to find out heat of enthalpy, we should use the equation $ {{\Delta H = \Delta U + W}} $ and modify W according to the conditions given in the question.