Thermodynamics - AP Chemistry

Card 0 of 1263

Question

Consider an electrochemical cell that has the following overall reaction:

2 H+(aq) + Sn (s) -> Sn2+ (aq) + H2 (aq)

Which of the following changes would alter the measured cell potential?

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

All of these changes would change Q and thus change the measured cell potential.

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Question

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq) + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions:

\[Al3+\] = 2.0 M; \[Mn2+\] = 1.0 M.

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n}: log: Q$

Altering these conditions would increase Q, and thus result in a decrease in the measured cell potential.

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Question

Determine the Ecell for the following reaction at 25 C:

Zn (s) + 2 VO2+ (aq) + 4 H+ -> 2 VO2+ (aq) + Zn2+(aq) + 2 H2O (l)

Given that:

VO2+ (aq) + 2 H+ (aq) + e- -> VO2+(aq) + H2O (l) Eo = 1.00 V

Zn2+ (aq) + 2 e--> Zn (s) Eo = -0.76 V

And

\[ VO2+\] = 2.0 M; \[H+\] = 0.50 M; \[VO2+\] = 1.0 x 10-2M; \[Zn2+\] = 1.0 x 10-1M

Answer

$E_{cell} = E^0_{cell} - \frac {0.0592}{n} : log : Q$

$E^o_{cell} = 1.00 - (-0.76) = 1.76 V$

$E_{cell} = 1.76 V - \frac{ 0.0592}{2} : log : \frac{[VO^{2+} ]^2 [Zn^{2+} ]}{[VO_2^+ ]^2 [H^+ ]^4 )}$

$= 1.76 V - \frac{0.0592}{2}: log : \frac{[1.0 x 10^{-2} ]^2 [1.0 x 10^{-1} ]}{[2.0]^2 [0.5]^4} = 1.89$

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Question

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq) + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions: \[Al3+\] = 1.0 M; \[Mn2+\] = 5.0 M.

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

At these concentrations Q will become smaller, and thus log Q will become smaller. This will give rise to a larger cell potential.

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Question

What is the cell potential of the following cell:

Zn (s) + 2 H+(aq) -> Zn2+ (aq) + H2 (g) Eo = 0.76 V

When the \[Zn2+\] = 1.0 M; PH2 = 1 atm, and the pH in the cathode is 5.2?

Answer

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log: Q$

$[H^+ ] = 10^{-pH}= 10^{-5.2} = 6.31 x 10^{-6}$

$E_{cell} = 0.76 V - \frac{0.0592}{2} : log : \frac{[Zn^{2+} ] P_{H_2} }{[H^+ ]^2 }$

$E_{cell} = 0.76 - \frac{0.0592}{2} : log: \frac{[1.0]1}{[6.31 x 10^{-6} ]^2} = 0.45 V$

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Question

Calculate the standard cell potential of the following reaction:

Cd(s) + MnO2 (s) + 4 H+ (aq) + -> Cd2+ (aq) + Mn2+ (aq) + 2 H2O (l)

Given:

MnO2 (s) + 4 H+ (aq) + 2e- -> Mn2+ (aq) + 2 H2O (l) Eo = 1.23 V

Cd2+ (aq) + 2 e- -> Cd (s) Eo = -0.40 V

Answer

Eocell = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

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Question

Calculate the standard cell potential of the following reaction:

Zn (s) + Cu2+ (aq) -> Zn2+ (aq) + Cu (s)

Given:

Zn2+(aq)+ 2 e--> Zn (s) Eo = -0.76 V

Cu2+(aq)+ 2 e--> Cu (s) Eo = 0.34 V

Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.34 – (-0.76) = 1.10 V

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Question

Calculate the standard cell potential of the following reaction:

Zn (s) + 2 Ag1+ (aq) -> Zn2+ (aq) + 2 Ag (s)

Given:

Zn2+(aq)+ 2 e--> Zn (s) Eo = -0.76 V

Ag1+(aq)+ 1 e--> Ag (s) Eo = 0.80 V

Answer

Eocell = Eo cathode - Eoanode

Eocell = 0.80 – (-0.76) = 1.56 V

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Question

Calculate the standard cell potential of the following reaction:

3 F2 (g) + 2 Au (s) -> 6 F- (aq) + 2 Au3+

Given:

F2 (g) + 2 e- -> 2 F- (aq) Eo = 2.87 V

Au3+(aq)+ 3 e--> Au (s) Eo = 1.50 V

Answer

Eocell = Eo cathode - Eoanode

Eocell = 2.87 – (1.50) = 1.37 V

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Question

H2O ice melts to liquid H2O \[Heat is added\]

Which of the following is correct?

I. Δ H is positive

II. ΔS is positive

III. ΔH is negative

IV. ΔS is negative

Answer

When heat is added, enthalpy increases, ΔH refers to enthalpy

ΔS refers to entropy, which is disorder; disorder increases as it moves from solid to liquid to gas state.

As ice goes to liquid water, entropy increases

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Question

Calculate ΔH for the following reaction:

CH4 (g) + O2 (g) ⇌ CO2 (g) + H2O (l)

Compound ΔH

CH4 (g) -74.8 kJ/mol

H2O (l) -285.8 kJ/mol

CO2 (g) -393.5 kJ/mol

Answer

ΔH = Σ ΔHf products - Σ ΔHf reactants

ΔHf O2 or any element is 0

First step is to balance the equation:

CH4 (g) + O2 (g) ⇌ CO2 (g) + 2H2O (l)

ΔH = Σ ΔHf products - Σ ΔHf reactants

= \[-393.5 kJ/mol + 2(-285.8) kJ/mol\] - (-74.8 kJ/mol)

= -890.3 kJ/mol

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Question

Which of the following is a true statement regarding the energy involved in the formation of Methane, CH4, from graphite C(s) and H2(g) ?

Answer

This problem requires a knowledge of the energy involved when bonds are broken versus when bonds are formed. Energy is released when bonds are formed, and energy is consumed when bonds are broken. Here, methane is being formed from C and H2. Since the carbon source, graphite, is monoatomic the carbon will be involved only in the formation of C–H bonds on it's way to forming methane. The H however exists as H2 so bonds will need to be broken to allow for C–H bonds to form. So first energy is consumed to break the H–H bonds, it is then released when the C–H bonds form.

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Question

The formation of nitrogen dioxide is a two step process.

$\small N_{2} + O_{2} \rightarrow 2NO$ $\small \Delta H = 180kJ$

$\small 2NO + O_{2}\rightarrow 2NO_{2}$ $\small \Delta H = -112kJ$

The net reaction is $\small N_{2} + 2O_{2} \rightarrow 2NO_{2}$ .

What is the change in enthalpy when creating four moles of nitrogen dioxide?

Answer

Hess's law states that the change in enthalpy for a total reaction can be considered equal to the sum of the enthalpy changes for every step involved in the reaction. In other words, we can determine the enthalpy change for nitrogen dioxide by adding the enthalpy changes for both steps involved in its formation.

$\Delta H_{step1}+\Delta H_{step2}=\Delta H_{}$

This gives us the total change in enthalpy for the listed reaction, $\small N_{2} + 2O_{2} \rightarrow 2NO_{2}$ . Because the question asks for the enthalpy change for four moles of nitrogen dioxide, the value must be doubled. The reaction only produces two moles of nitrogen dioxide.

$\frac{68kJ}{2mol\ NO_2}*2=\frac{136kJ}{4mol\ NO_2}$

$\Delta H_{4mol}=136kJ$

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Question

Which of the following statements is true concerning a chemical reaction?

Answer

When a chemical reaction is represented graphically, we see that the enthalpy change is reversed between the forward and reverse reactions. If a reaction produces energy in a forward process, it will require an input of energy in the reverse process, and vice versa.

$reactant\ +heat\rightleftharpoons product,\ \Delta H=+$

$product\rightleftharpoons heat\ +reactant,\ \Delta H=-$

A catalyst only affects the rate of a chemical reaction; it does not affect the equilibrium. Finally, exothermic reactions are not always spontaneous, but will have lower activation of energies compared to endothermic reactions.

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Question

$A \rightarrow E+F$ $\Delta H=-50kJ$

$E+F\rightarrow C+G$ $\Delta H=+85 kJ$

$B+G \rightarrow D$ $\Delta H= -10kJ$

What is the change in enthalpy for the given reaction?

$A+B\rightarrow C+D$

Answer

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known reactions. In this case the known reactions are given.

$A \rightarrow E+F$ $\Delta H=-50kJ$

$E+F\rightarrow C+G$ $\Delta H=+85 kJ$

$B+G \rightarrow D$ $\Delta H= -10kJ$

Since the reactions are in the correct order, adding all the $\Delta H$ values together can be used to calculate the $\Delta H$ of the reaction.

$A+E+F+B+G\rightarrow E+F+C+G+D$

$\text{Net reaction: }A+B\rightarrow C+D$

$\Delta H=-50kJ+85kJ-10kJ=25kJ$

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Question

$N_{2(g)}$ $\Delta H_f=0 kJ$

$O_{2(g)}$ $\Delta H_f=0 kJ$

$\Delta H_f=90kJ$

What is the enthalpy of the following reaction?

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

Answer

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$N_{2(g)}$ $\Delta H_f=0 kJ$

$O_{2(g)}$ $\Delta H_f=0 kJ$

$\Delta H_f=90kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction. For this particular reaction, since there are two moles of product, the enthalpy of formation for must be multiplied by two.

$N_{2(g)}+O_{2(g)}\rightarrow 2NO$

$\Delta H_{rxn}=2(\Delta H_{NO})-\Delta H_{N_{2(g)}}-\Delta H_{O_{2(g)}}$

$\Delta H_{rxn}=2(90kJ)-0kJ-0kJ$

$\Delta H_{rxn}=180\frac{kJ}{mol}$

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Question

$NO_{(g)}$ $\Delta H_f=90kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$NO_{2}$ $\Delta H_f=33kJ$

What is the change in enthalpy for the following reaction?

$NO_{(g)}+O_{2(g)} \rightarrow NO_2$

Answer

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$NO_{(g)}$ $\Delta H_f=90kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$NO_{2}$ $\Delta H_f=33kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$2NO_{(g)}+O_{2(g)}\rightarrow 2NO_{2}$

$\Delta H_{rxn}=2(\Delta H_{NO_2})-2(\Delta H_{NO_{(g)}})-\Delta H_{O_{2(g)}}$

$\Delta H_{rxn}=2(33kJ)-2(90kJ)-0kJ$

$\Delta H_{rxn}=-114\frac{kJ}{mol}$

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Question

$\Delta H_f=33kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$N_{2(g)}$ $\Delta H_f=0kJ$

What is the change in enthalpy for the following reaction?

$N_{2(g)}+O_{2(g)}\rightarrow NO_2$

Answer

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$\Delta H_f=33kJ$

$O_{2(g)}$ $\Delta H_f=0kJ$

$N_{2(g)}$ $\Delta H_f=0kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$N_{2(g)}+2O_{2(g)}\rightarrow 2NO_2$

$\Delta H_{rxn}=2(\Delta H_{NO_2})-\Delta H_{N_{(g)}}-2(\Delta H_{O_{2(g)}})$

$\Delta H_{rxn}=2(33kJ)-0kJ-2(0kJ)$

$\Delta H_{rxn}=66\frac{kJ}{mol}$

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Question

$\Delta H_f=0kJ$

$\Delta H_f=0kJ$

$\Delta H_f=-601kJ$

What is the change in enthalpy for the following reaction?

$2Mg +O_2\rightarrow 2MgO$

Answer

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$\Delta H_f=0kJ$

$\Delta H_f=0kJ$

$\Delta H_f=-601kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$2Mg +O_2\rightarrow 2MgO$

$\Delta H_{rxn}=2(\Delta H_{MgO})-2(\Delta H_{Mg})-\Delta H_{O_{2}}$

$\Delta H_{rxn}=2(-601kJ)-2(0kJ)-0kJ$

$\Delta H_{rxn}=-1202\frac{kJ}{mol}$

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Question

$C_4H_{10}$ $\Delta H_f=-126kJ$

$\Delta H_f=0kJ$

$\Delta H_f=-394kJ$

$\Delta H_f=-286kJ$

What is the change in enthalpy for the following reaction?

$C_4H_{10}+O_2\rightarrow CO_2+H_2O$

Answer

The change in enthalpy is calculated by:

$\Delta H_{products} - \Delta H_{reactants}$

When $\Delta H$ cannot be measured, it can be calculated from known enthalpies of formation.

$C_4H_{10}$ $\Delta H_f=-126kJ$

$\Delta H_f=0kJ$

$\Delta H_f=-394kJ$

$\Delta H_f=-286kJ$

It is important to first balance the reaction before performing calculations. The coefficients are important in determining the change in enthalpy of a reaction.

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

$\Delta H_{rxn}=8(\Delta H_{CO_2})+10(\Delta H_{H_2O})-2(\Delta H_{C_4H_{10}})-13(\Delta H_{O_2})$

$\Delta H_{rxn}=8(-394kJ)+10(-286kJ)-2(-126kJ)-13(0kJ)$

$\Delta H_{rxn}=-5760\frac{kJ}{mol}$

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