The deltaHrxn = -243 kJ/mol the deltaHrxn of CH4(methane) = -802 kJ/mol
The fuel that yields more energy per mole is METHANE. The negative sign merely signifies the release of energy. Thus, 802 kJ/mol is greater than 243 kJ/mol.
The fuel that yields more energy per gram is HYDROGEN. Here is the computation:
deltaHrxn = (-243 kJ/mol)(1 mol/2.016 g H2) <span>= -120.535714286 kJ/g or -121 kJ/g
</span>deltaHrxn of CH4(methane) = (-802 kJ/mol)(1 mol/16.04 g)
<span>= -50 kJ/g
</span>
As discussed the negative sign serves as the symbol of released energy. Thus, 121 is greater than 50.
Answer:
Option C . CO2(g) + H2O(g)
Explanation:
When hydrocarbon undergoes combustion, carbon dioxide (CO2) and water (H2O) are produced.
C2H4(g) + O2(g) —› CO2(g) + H2O(g)
Thus, the product of the unbalanced combustion reaction is:
CO2(g) + H2O(g)
Thus, we can balance the equation as follow:
C2H4(g) + O2(g) —› CO2(g) + H2O(g)
There are 2 atoms of C on the left side and 1 atom on the right side. It can be balanced by putting 2 in front of CO2 as shown below:
C2H4(g) + O2(g) —› 2CO2(g) + H2O(g)
There are 4 atoms of H on the left side and 2 atoms on the right side. It can be balanced by putting 2 in front of H2O as shown below:
C2H4(g) + O2(g) —› 2CO2(g) + 2H2O(g)
There are a total of 6 atoms of O on the right side and 2 atom on the left side. It can be balanced by putting 3 in front of O2 as shown below:
C2H4(g) + 3O2(g) —› 2CO2(g) + 2H2O(g)
Thus, the equation is balanced.
Answer:
A catalyst provides an alternative pathway for the reaction, that has a lower activation energy. ... Increasing the temperature will mean there are more particles with high energy, greater than the activation energy, so a higher chance of a successful reaction.
It would be a combustion reaction (B) because the products are carbon dioxide and water.
