Question

1. a reaction will never be spontaneous when

a) δg°>0.
b) δh°<0 and δs°>0.
c) δh°<0 and δs°<0.
d) δg=0.
e) δg>0.

1. the copper (i) ion is a curious species. in aqueous solutions, there are a number of reactions that it can undergo; one is the reaction with other copper (i) ions:

2cu⁺(aq)⇌cu²⁺(aq)+cu(s)
using the tabulated data, calculate the equilibrium constant for this reaction of cu⁺(aq) at 298 and predict whether it will increase or decrease with increasing temperature. choose from the following.
a) 3.5×10⁴; increase
b) 1.2×10⁶; decrease
c) 2.2×10⁷; increase
d) 2.2×10⁷; decrease
e) 3.5×10⁴; decrease

1. consider the aqueous phase reaction between the dichromate anion and iron (ii) cations:

14h₃o⁺(aq)+cr₂o₇²⁻+6fe²⁺(aq)→2cr³⁺(aq)+21h₂o
what is the reaction rate expressed in terms of changing h₃o⁺ concentration?
a) reaction rate = 14δh₃o⁺/δt
b) reaction rate = -14δh₃o⁺/δt
c) reaction rate = -1/14δh₃o⁺/δt
d) reaction rate = 1/14δh₃o⁺/δt
e) reaction rate = -δh₃o⁺/δt

Explanation:

Step1: Calculate $\Delta H^{\circ}$

Using the formula $\Delta H^{\circ}=\sum n\Delta H_{f}^{\circ}(products)-\sum m\Delta H_{f}^{\circ}(reactants)$
$$\Delta H^{\circ}=(64.77 + 0)-2\times71.67=-78.57\space kJ/mol$$

Step2: Calculate $\Delta S^{\circ}$

Using the formula $\Delta S^{\circ}=\sum nS^{\circ}(products)-\sum mS^{\circ}(reactants)$
$$\Delta S^{\circ}=(-99.6 + 33.15)-2\times40.6=-147.65\space J/(mol\cdot K)$$

Step3: Calculate $\Delta G^{\circ}$ at $T = 298K$

Using the formula $\Delta G^{\circ}=\Delta H^{\circ}-T\Delta S^{\circ}$
First convert units: $\Delta H^{\circ}=-78570\space J/mol$
$$\Delta G^{\circ}=-78570-298\times(-147.65)=-78570 + 44009.7=-34560.3\space J/mol$$

Step4: Calculate the equilibrium constant $K$

Using the formula $\Delta G^{\circ}=-RT\ln K$ ($R = 8.314\space J/(mol\cdot K)$)
$$\ln K=\frac{-\Delta G^{\circ}}{RT}=\frac{34560.3}{8.314\times298}\approx14$$
$$K = e^{14}\approx2.2\times10^{6}$$

Step5: Determine the effect of temperature on $K$

Using the van't Hoff equation $\frac{d\ln K}{dT}=\frac{\Delta H^{\circ}}{RT^{2}}$
Since $\Delta H^{\circ}<0$, as $T$ increases, $\ln K$ decreases, so $K$ decreases.

Answer:

b) $1.2\times10^{6}$; decrease