Chem 1050 Midterms

Each of the following objects were heated individually from 22 C to 41 C. Which required the most energy?

• 15 g of H2O (Cₛ = 4.18 J g⁻¹ C⁻¹)

• 60 g of Al (Cₛ = 0.902 J g⁻¹ C⁻¹)

• 20 g of Au (Cₛ = 0.128 J g⁻¹ C⁻¹)

• 10 g of Cu (Cₛ = 0.385 J g⁻¹ C⁻¹)

• 80 g of CCl₄ (Cₛ = 0.861 J g⁻¹ C⁻¹)

Answer explanation

We are given the specific heat capacity and the mass of the substance. Multiply them
together to get J/ ̊C. Since they all experience the same temperature increase, the one that
requires the largest number of joules to heat it per degree will take the most energy overall.
15 x 4.18 = 62.7 J/ ̊C
60 x 0.902 = 54.12 J/ ̊C
20 x 0.128 = 2.56 J/ ̊C
10 x 0.385 = 3.85 J/ ̊C
80 x 0.861 = 68.88 J/ ̊C
Clearly the answer is E, CCl 4

The molar enthalpy of combustion of sucrose (molar mass is 342 g mol⁻¹) is -5650 kJ mol⁻¹. What mass of sucrose has been burned when 210 kJ of heat is produced by its combustion?

• 3.72 g

• 16.5 g

• 9.20 g

• 12.7 g

• 26.9 g

Answer explanation

Shown is a graph of temperature vs. time for the heating of a substance we will call X. The rate of energy being supplied is constant. Which phase of X has the largest molar heat capacity?

• X(s) solid.

• Solid and liquid are about the same.

• X(l) liquid.

• Liquid and gas are about the same.

• X(g) gas.

Answer explanation

With the melting temperature identified, we know
this graph must start with X in the solid state. The
first rising line is heating the solid to its melting
point. The slope of the line is its heat capacity. When
the temperature rises quickly, the heat capacity
must be small. The flat region at the melting point is
due to the enthalpy of melting. The next rise is to the
vapour phase, so it is the heating of the liquid phase. Its slope is smaller than that for the
heating the solid, so it takes more heat to raise the temperature a degree, so it has a larger
heat capacity. The next flat region is the enthalpy of vapourization. Finally the last rise is
due to the heat capacity of the gas phase X. Clearly the liquid heats the slowest and so it
must have the largest heat capacity.
The answer is C

When ammonia, NH3, evaporates at constant pressure, the sign of the enthalpy change for the process

• is negative

• is positive

• depends upon the external pressure

• depends upon the temperature

• cannot be predicted with this information as it can only be measured experimentally

Answer explanation

Evapouration always requires the input of energy so the liquid molecules can break their
bonds with each other and escape into the gas phase. So energy going intoa system is
defined as being positive, the answer must B

An ideal gas expands isothermally (at constant temperature). In an isothermal process, the internal energy of an ideal gas is unchanged. From the perspective of the system, the heat exchanged in the process is +85 J. What is value for the work?

• -85 J

• +85 J

• 0 J

• +170 J

• -170 J

Answer explanation

This is just a First Law of Thermodynamics problem. The question tells us the ∆U q + w = 0. So if the heat is +85 J, then the work must equal -85 J. The answer is A

The reaction to form glutamine from glutamate and ammonia is glutamate + NH3 → glutamine ΔG°' = +14.1 kJ mol^-1 (1) There is an enzyme that can couple this reaction to the ATP/ADP reaction: ATP → ADP+ PO4^3-(aq) ΔG°' = -30.5 kJ mol^-1 (2) Coupling these two reactions favours the formation of glutamine because.

• Reaction 1 favours reactants.

• Reaction 2 favours reactants.

• Reaction 2 is more exergonic than reaction 1 is endergonic.

• The sum of reactions 1 and 2 give a reaction which favours reactants.

• The sum of reaction 2 with the reverse of reaction 1 gives a reaction with a negative ΔG.

Answer explanation

Answer A, while true because of its positive Gibbs energy, that alone does not favour the production of glutamine. Answer B, with its negative Gibbs energy, actually favour products. In answer C, reaction 2 is indeed more exergonic (releases Gibbs energy) than reaction 1 is endergonic (it absorbs Gibbs energy). Overall the process has a negative ∆G˚ and is spontaneous. This is process by which glutamine is formed. For D, the sum of 1 and 2 favours products, not reactants. For E, the stated sum does produce a negative ∆G˚, but it consumes glutamine, not produces it. The answer is C

Which expression describes the heat evolved in a chemical reaction that is carried out at constant pressure?

• ΔU – w

• q + w

• qH - ΔU

• ΔU – q

• ΔU + w

Answer explanation

Look to the First Law to answer this question. ∆U = q + w is always true. Rearrange it to read q = ∆U – w. We see right away that the answer must be A.

A model gas phase reaction, A(g) → B(g), has a Gibbs energy, ΔG, equal to 0 kJ mol^-1 under certain conditions. At the same time and under those same conditions, ΔG° = +30 kJ mol^-1. Which statement is true about this reaction at these conditions?

• The reaction will proceed in the forward direction to produce more B.

• The reaction will proceed in the reverse direction to produce more A.

• The reaction is at equilibrium and [A] > [B].

• The reaction is at equilibrium and [A] = [B].

• The reaction is at equilibrium and [A] < [B].

Answer explanation

∆G˚ tells us about the location of equilibrium for a reaction. ∆G tells us in which direction we need to move to reach equilibrium. In this case, ∆G = 0 so that means we are at equilibrium. We have to choose between C, D, and E. Recall that ∆G˚ = -RT lnK and K will have the form of [B]/[A]. For ∆G˚ to be positive, lnK must be negative. That means that K < 1. And that means that [A] > [B]. The answer, therefore, is C.

Imagine two hypothetical elements, X and Y. At room temperature, the standard state for both is as a diatomic gas (X2 and Y2). They can react to form the gas XY, whose standard enthalpy of formation is -150 kJ mol^-1. The bond enthalpies for the two elements are BE(X2) = +200 kJ mol^-1 and BE(Y2) = +100 kJ mol^-1. What would be a correct estimate of the bond enthalpy of the X-Y bond?

• +200 kJ mol^-1

• +250 kJ mol^-1

• +300 kJ mol^-1

• +350 kJ mol^-1

• -400 kJ mol^-1

Answer explanation

The production of quicklime (calcium oxide) comes from heating calcium carbonate according to this reaction equation. CaCO3(s) → CaO(s) + CO2(g) With the values of ΔH° = +179 kJ mol^-1 and ΔS° = +160 J K^-1 mol^-1, under which temperature conditions will the equilibrium constant take on a value K < 1? At no temperature.

• At all temperatures.

• At temperatures below 1118 K.

• At temperatures above 1120 K.

• At temperatures greater than 500 C.

Answer explanation

K>1 correlates with ∆G˚ becoming negative, based on the equation ∆G˚ = -RT lnK. ∆G˚ = ∆H˚ - T∆S˚ = 179,000 – T(160) = 0. Solving for T we get T = 179,000/160 = 1118.75 K The statement in the question that yu have to heat calcium carbonate would suggest that it is above this temperature when K becomes positive. You could also see this by looking at the equation. If T is smaller, the T∆S˚ term will be smaller and less negative so that ∆G˚ must be positive. Therefore, of the answers provided, only D is true.