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Pad 1:44 PM D UVU Login Servic × 'MyLab and Mast: × ' D Cours MasteringChemis × https/session.masteringchemistry.com/myct/itemvie.. twe10 ± Entapy of Reaction State and Stoichiometry previous 12 of 24 |next s Hints Reset Help Same Diferent My Answers Give Up Part B This question will be shown after you complete previous question(s) Part C This question wil be shown after you complete previous question(s) Part D Suppose that 0.370 mol of methane, CH4g) is reacted with 0.520 mol of fluorine, Fa(g).forming CF4(g) and HF(g) as sole products. Assuming that the reaction occurs at constant pressure, how much heat is released? Express your answer to three significant figures and include the appropriate units heat-Value Units My Answers Give UpExplanation / Answer
Answer:
The balanced equation is
CH4 + 4F2-----> CF4 + 4HF
Given 0.370 mol CH4 and 0.520 mol of fluorine F2
From the equation the mole ratio between 2 is 1:4
So moles of F2=4xmol CH4=4x0.370 mol=1.48 mol F2.
But here we have only 0.520 mol F2. So limiting reacgent is F2.
Now converting 0.520 mol F2 to moles of CF4.
0.520 mol F2 x (1 mol CF4/4 mol F2) = 0.520 x 1/4 = 0.13 mol CF4.
Next we must determine how much of the CH4 reacted and how much HF is produced.
0.52 mol F2 x (1 mol CH4/4 mol F2) = 0.520 x 1/4 = 0.13 mol CH4 used.
We know 0.13 mol CF4 is produced; therefore, we must have produced 4 x 0.130 = 0.52 mol HF.
Now we plug data into the delta H rxn using delta Ho formation values that you probably can find in your text.
dHrxn = sum of (mols*dH products) - sum of (mols *dH reactants).
dHrxn is the heat produced by that reaction.
Given the values of enthalpy are
CH4 = -74.8kJ/mole
CF4 = -925.0 kJ/mole
HF = -271.1 kJ/mole
F2=0
dHrxn = (0.13 x -925kJ/mole+0.52 x -271.1kJ/mole ) - (0.13 x -74.8kJ/mole+0.52x0) =-251.498 kJ/mol.
Therefore dHrxn = -251.498 kJ/mol.
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