Question

For the reaction, ${{C}_{2}}{{H}_{4\left( g \right)}}+3{{O}_{2}}\to 2C{{O}_{2\left( g \right)}}+2{{H}_{2}}{{O}_{\left( l \right)}}$ ; $\Delta U=-1415kJ.$ Then $\Delta H$ at ${{27}^{o}}C$ is?

Answer 1

As we know that change in enthalpy of a reaction is the amount of heat released or being absorbed as the reaction takes place. We can calculate the change in enthalpy in different ways depending on the specific situation.

Complete answer:
We will discuss the simplest way of calculating the enthalpy change which uses the enthalpy of reactants and products. As we know that if we know these quantities then we can use the formula $\Delta H={{H}_{products}}-{{H}_{reac\tan ts}}$ to find the overall change.
If we consider the reaction of Hydrogen ion with oxygen ion to form a hydrogen oxygen compound, then we can calculate the enthalpy change in this way. As we know that oxygen ion has an enthalpy of - $-1415kJ$ and now, we will all these values given in the formula of enthalpy change as:
$\Delta H=\Delta E+\Delta ngRT$
$\Rightarrow \Delta H=\left[ \left( -1415 \right) \right]+\left[ \left( -2 \right)\times \left( 8.314 \right)\times \left( 300 \right) \right]$
$\therefore \Delta H=-1420KJ$
Hence, in this way we can calculate the enthalpy change.

Note:
As we know that enthalpy is an energy-like property that has dimensions of energy. It is mainly measured in joules or ergs. It is found that there are several factors that affect the enthalpy of reaction, these are the temperature of the system, and the concentration of the reactants and the partial pressure of the gases if any are involved.