Question

1) Express your answer as a molecular formula.

a) Use the data below to calculate the heat of hydration of lithium chloride.

b) Calculate the heat of hydration of sodium chloride.

Compound	Lattice Energy (kJ/mol)	ΔHsoln(kJ/mol)
LiCl	-834	-37.0
NaCl	-769	+3.88

2)A certain reaction with an activation energy of 115 kJ/mol was run at 485 K and again at 505 K . What is the ratio of f at the higher temperature to f at the lower temperature? Express your answer numerically using one significant figure. f505/f485=?

3)The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.30 kJ/mol and a frequency factor of A = 6.00×10⁸M−1⋅s−1 . The reaction is believed to be bimolecular:

NO(g)+F2(g)→NOF(g)+F(g)

What is the rate constant at 657 ∘C ?

4)Consider the following reaction:
2HBr(g)→H2(g)+Br2(g)

a)In the first 25.0 s of this reaction, the concentration of HBr dropped from 0.600 M to 0.458 M . Calculate the average rate of the reaction in this time interval.

b)If the volume of the reaction vessel in part (b) was 1.50 L, what amount of Br2 (in moles) was formed during the first 15.0 s of the reaction?

5)Consider the following reaction:
2N2O(g)→2N2(g)+O2(g)

a)In the first 14.0 s of the reaction, 1.7×10⁻² mol of O2 is produced in a reaction vessel with a volume of 0.360 L . What is the average rate of the reaction over this time interval?

b)Predict the rate of change in the concentration of N2O over this time interval. In other words, what is Δ[N2O]/Δt?

Answer 1

1) solution : the reation will be

a) Li⁺ + Cl ^- --> LiCl(s)   deltaH lattice =-834KJ /mole   .........equation 1

LiCl (s) + aq --> LiCl(aq) Delta H solution = -37 J / mole......equation 2

Li⁺ + Cl ^- --> LiCl (aq) Hydration

So Delta H hydration = Equation 1 + equation 2 = -834 + (-37 ) =-871 KJ / mol

b) Na⁺ + Cl ^- --> NaCl(s)   deltaH lattice = -769KJ /mole   .........equation 1

NaCl (s) + aq --> NaCl(aq) Delta H solution = 3.88J / mole......equation 2

Na⁺ + Cl ^- --> NaCl (aq) Hydration

So Delta H hydration = Equation 1 + equation 2 = - 769 + (+3.88 ) = - 765.12KJ / mol

2.

3.

Ea = 6.30 kJ/mol = 6.30X10³ J/mol

A = 6.00×10⁸ M⁻¹⋅s⁻¹

Temperature, T = 657^oC = 657 + 273

= 930 K

Rate constant ids given by the arhenious equation as:

K = A*e^{(- Ea/RT)}

K =  6.00X10⁸ *e^{(- 6300 /8.314*630)}

K = 6.00X10⁸ *e^{- 0.814793}

K = 6.00X10⁸ *0.442730

K = 2.6564X10⁸ M^-1 s^-1