what is the delta H of N2 + 3H2 ---> 2NH3 + 22,000 cal
To calculate the enthalpy change (ΔH) of a reaction, you need to know the enthalpy change of each individual molecule involved in the reaction. Fortunately, enthalpy changes are often provided in the form of a table.
In this case, we are given the enthalpy change in calories (cal) for the reaction:
N2 + 3H2 --> 2NH3
ΔH = -22,000 cal
The negative sign indicates that the reaction is exothermic, meaning it releases heat.
The enthalpy change, ΔH, represents the difference between the enthalpy of the products and the reactants. For this reaction, since we have the ΔH value, we can conclude that the enthalpy of the reactants is higher than the enthalpy of the products by 22,000 cal.
Therefore, the enthalpy change (ΔH) of the reaction N2 + 3H2 -> 2NH3 is -22,000 cal.
To find the ΔH (change in enthalpy) of a reaction, you need to calculate the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants.
The balanced chemical equation for the reaction is:
N2 + 3H2 ---> 2NH3
First, let's determine the enthalpies of the reactants and products using available information. The enthalpy of formation (∆Hf) of a compound is the amount of heat absorbed or released when 1 mole of the compound is formed from its constituent elements in their standard states. The ∆Hf values are usually given in kJ/mol.
Given:
∆Hf(N2) = 0 kJ/mol
∆Hf(H2) = 0 kJ/mol
∆Hf(NH3) = -46 kJ/mol
Next, we need to convert the given ΔH value from calories to joules by using the conversion factor:
1 cal = 4.184 J
Given:
ΔH = 22,000 cal
ΔH = 22,000 cal * 4.184 J/cal
Now we can calculate the ΔH for the reaction:
ΔH = (∑∆Hf(products)) - (∑∆Hf(reactants))
ΔH = (2 * ∆Hf(NH3)) - ((∆Hf(N2) + 3 * ∆Hf(H2)))
Substituting the values:
ΔH = (2 * -46 kJ/mol) - ((0 kJ/mol) + (3 * 0 kJ/mol))
Finally, convert the ΔH to the desired unit:
ΔH = (2 * -46 kJ/mol) * 1000 J/kJ
Thus, the value for ΔH of the reaction is -92,000 J.