Solid phosphorous is found in two forms: white phosphorous (P4) and red phosphorous (P). White phosphorous is the standard state. The enthalpy of the formation of red phosphorus is -17.6kJ.
When 32.6 g of white phosphorous reacts to form red phosphorous. What is the enthalpy change?
Note the correct spelling of phosphorus. The word you spelled is a word but not the correct one for the element, either P4 or P. For example, H3PO3 is phosphorous acid.
P4 ==> 4P + 17.6 kJ.
17.6 kJ x (32.6/3*molar mass P4) = ? and it is exothermic which makes the sign for delta H negative.
To find the enthalpy change when 32.6 g of white phosphorus reacts to form red phosphorus, we need to use the concept of stoichiometry, which relates the amount of reactant and product in a chemical reaction.
First, we calculate the number of moles of white phosphorus. We can use the molar mass of white phosphorus (P4) to do this. The molar mass of white phosphorus is 4 * atomic mass of phosphorus, which is approximately 4 * 31 g/mol = 124 g/mol.
Number of moles of white phosphorous = Mass of white phosphorus / Molar mass of white phosphorus
= 32.6 g / 124 g/mol
Now, we have the number of moles of white phosphorus. Next, we need to use the balanced chemical equation for the reaction between white phosphorus and red phosphorus to determine the stoichiometric ratio between them.
The balanced equation for the reaction is:
P4 (white phosphorus) -> 4P (red phosphorus)
From the equation, we can see that 1 mole of white phosphorus reacts to form 4 moles of red phosphorus.
Now, we can calculate the number of moles of red phosphorus formed:
Number of moles of red phosphorus = Number of moles of white phosphorus * (4 moles of red phosphorus / 1 mole of white phosphorus)
Once we have the number of moles of red phosphorus formed, we can use the enthalpy change of formation of red phosphorus (-17.6 kJ) to determine the enthalpy change for the given mass of white phosphorus.
Enthalpy change = Number of moles of red phosphorus * Enthalpy change of formation of red phosphorus
Calculate these values to get the final answer.