1.) For which of the following substances is the standard enthalpy of formation equal to zero?
a) water [H2O(l)] d) carbon dioxide [CO2(g)]
b) lead [Pb(l)] e) tin [Sn(s)]
c) carbon dioxide [CO2(s)]
2.) Consider the following four equations:
1. C6H6(l) + O2(g) 6CO2(g) + 3H2O(l) ΔH1
2. H2(g) + ½ O2(g) H2O(l) ΔH2
3. C(s) + O2(g) CO2(g) ΔH3
4. 6C(s) + 3H2(g) C6H6(l) ΔH4
The enthalpy change for reaction 1, ΔH1, can be obtained by combining the ΔH’s for the other reactions in which of the following ways?
a) 6(ΔH3) + 3(ΔH2) + ΔH4
b) 6(ΔH3) - 3(ΔH2) + ΔH4
c) 6(ΔH3) - 3(ΔH2) - ΔH4
d) 6(ΔH3) + 3(ΔH2) - ΔH4
e) none of these
3.) Which forces exist between hydrogen chloride particles?
I. London forces II. Metallic bonding III. Hydrogen bonding IV. Dipole-dipole
a) I only d) I, III and IV only
b) I and IV only e) I, II and III only
c) I and II only
1.) To determine which substance has a standard enthalpy of formation equal to zero, you need to know the definition of standard enthalpy of formation. The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states.
To find the answer, you need to know the standard enthalpies of formation for the substances listed. Look up the standard enthalpy of formation values for water (H2O), carbon dioxide (CO2), lead (Pb), and tin (Sn) in their respective states (liquid l or solid s) and compare them to zero. Whichever substance has a standard enthalpy of formation equal to zero is the answer.
2.) To find the enthalpy change (ΔH1) for reaction 1, you need to combine the enthalpies (ΔH) of the other reactions in a way that cancels out the reactants and products that appear on both sides of the equation.
By looking at the reactions, you can see that reaction 3 (C + O2 → CO2) produces the same amount of CO2 as reaction 1, but with different stoichiometry. So, you can use reaction 3 to cancel out the CO2 term in reaction 1.
To balance the number of CO2 molecules, you need to multiply reaction 3 by 6. To balance the number of H2O molecules, you need to multiply reaction 2 by 3. Finally, reaction 4 (6C + 3H2 → C6H6) needs to be reversed to match the stoichiometry of reaction 1.
Adding all of these reactions together will give you the overall reaction with the correct stoichiometry: 6(ΔH3) + 3(ΔH2) - ΔH4. Therefore, the correct answer is d) 6(ΔH3) + 3(ΔH2) - ΔH4.
3.) To determine which forces exist between hydrogen chloride (HCl) particles, you need to understand the types of intermolecular forces.
London forces, also known as van der Waals forces, are present between all particles and arise from temporary fluctuations in electron density. They are the weakest type of intermolecular force.
Metallic bonding occurs in metals and involves the sharing of delocalized electrons among metal atoms.
Hydrogen bonding is a special type of dipole-dipole interaction that occurs when a hydrogen atom is bonded to a highly electronegative atom (such as oxygen or nitrogen) and is attracted to another electronegative atom nearby.
Dipole-dipole forces occur between polar molecules, where the positive end of one molecule is attracted to the negative end of another molecule.
Based on these definitions, hydrogen chloride (HCl) molecules have dipole-dipole forces and London forces. Therefore, the correct answer is b) I and IV only (dipole-dipole forces and London forces).
OK. I've answered two questions that contained 3 problems each. I think you;ve had a chance to read my note to the first one by now. Here goes #1 ONLY.
Elements in their pure state at 25 C have DH = zero and 1 atm pressure are zero.