Question

The difference in $\Delta H$ and $\Delta E$ for the combustion of methane at $27^\circ {\rm{C}}$ in calories would be:
A.1800
B.162
C.1200
D.160

Answer 1

Hint: We know that combustion is the reaction in which a fuel is burnt in presence of oxygen to give carbon dioxide and water. Here, we have to use the formula $\Delta H = \Delta E + \Delta {n_g}RT$ to calculate the difference in $\Delta H$ and $\Delta E$. In the formula, $\Delta H$ represents change in enthalpy, $\Delta E$ is change in internal energy, $\Delta {n_g}$ is change in moles (gaseous), R is gas constant and T is temperature.

Complete step by step answer:
Let’s first understand fuel in detail. A fuel is a hydrocarbon such as methane, ethane, butane etc. Fuel undergoes combustion in presence of oxygen to produce water and carbon dioxide. Let’s first understand fuel in detail. A fuel is a hydrocarbon such as methane, ethane, butane etc. Fuel undergoes combustion in presence of oxygen to produce water and carbon dioxide. General term used for combustion is burning.

Now, we write the combustion reaction for methane.

${\rm{C}}{{\rm{H}}_{\rm{4}}}\left( g \right) + 2{{\rm{O}}_{\rm{2}}}\left( g \right) \to {\rm{C}}{{\rm{O}}_{\rm{2}}}\left( g \right) + 2{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right)$… (1)

Now, we have to calculate the change in number of moles. This can be calculated by subtracting moles of reactant from moles of product. The formula is,

$\Delta {n_g} = {n_{product}} - {n_{reac\tan t}}$ ….. (2)

Now, we have to use equation (1) to know the moles of the product and reactant. Only for gaseous products or reactants number of moles is counted. So, moles of products is 1 and moles of reactants is 3. Now, we put these values in equation (2).

$ \Rightarrow \Delta {n_g} = 1 - 3 = - 2$

The temperature is given as $27^\circ {\rm{C}}$. So, we have to convert it into Kelvin.

$ \Rightarrow 27^\circ {\rm{C}} = {\rm{27 + 273}} = {\rm{300}}\,{\rm{K}}$

The value of gas constant (R) is 2 Calorie.

Now, we have to put the values of R, T and $\Delta {n_g}$ in the below formula.

$\Delta H = \Delta E + \Delta {n_g}RT$

$ \Rightarrow \Delta H - \Delta E = {n_g}RT$

$ \Rightarrow \Delta H - \Delta E = - 2 \times 2 \times 300\,$

$ \Rightarrow \Delta H - \Delta E = - 1200\,{\rm{cal}}$

Therefore, in the combination of methane, the difference in $\Delta H$ and $\Delta E$ is -1200 calories.

Therefore, the correct option is C. .

Note: Always remember that, the first law of thermodynamics states that change in internal energy is equal to the summation of heat added to the system and work done on the system.
$\Delta U = q + w$
Here,$\Delta U$ represents change in internal energy, q is heat added to the system and w is work done on the system.
In case work is done by the system, the equation becomes,
$\Delta U = q - w$