Answer the following questions based on 2NH3->N2+3H2+450Joules
How much heat is released in the equation: 6NH3->3N2+9H2
What is the heat of the reaction: H2+1/3N2->2/3NH3
Equation 2 is 3x that of 1; therefore, the heat released is 3x450 J.
Equation 3 is 1/3 of the reversed equation 1; therefore, the heat absorbed is 1/3 x 450J.
To determine the heat released or absorbed in a chemical reaction, we use the concept of the heat of reaction, also known as the enthalpy change (∆H) of the reaction. The enthalpy change represents the amount of heat energy either released or absorbed during a chemical reaction.
For the given equation 2NH3 → N2 + 3H2 + 450 Joules, the heat energy of 450 Joules is specified. This means that 450 Joules of heat are released during the reaction.
1. How much heat is released in the equation: 6NH3 → 3N2 + 9H2?
To find the heat released in a different equation, you can use the concept of stoichiometry. Stoichiometry is a way to balance chemical equations and determine the mole ratios between reactants and products.
In the given equation, which is a multiple of the original equation, the stoichiometric coefficients (numbers in front of the compounds) are doubled compared to the original equation. This means that the heat released will also double.
Therefore, if the original equation releases 450 Joules of heat, the equation 6NH3 → 3N2 + 9H2 will release 2 * 450 = 900 Joules of heat.
2. What is the heat of the reaction H2 + 1/3N2 → 2/3NH3?
To determine the heat of a different reaction, you will need experimental data or thermodynamic values, specifically the heat of formation (∆Hf) of the compounds involved. This value represents the change in enthalpy when one mole of a compound forms from its constituent elements in their standard states.
Considering the given reaction, H2 + 1/3N2 → 2/3NH3, you will need the ∆Hf values of H2, N2, and NH3.
The ∆Hf values can be found in thermodynamic tables or databases. For example, let's assume the ∆Hf values are as follows:
∆Hf(H2) = -286 kJ/mol
∆Hf(N2) = 0 kJ/mol
∆Hf(NH3) = -46 kJ/mol
To calculate the heat of the reaction, add the ∆Hf values of the products and subtract the sum of the reactant's ∆Hf values:
∆H = (∆Hf(N2) + 3∆Hf(H2)) - (∆Hf(NH3))
= (0 + 3(-286)) - (-46)
= -858 - (-46)
= -858 + 46
= -812 kJ/mol (heat of the reaction)
Therefore, the heat of the reaction H2 + 1/3N2 → 2/3NH3 is -812 kJ/mol.