Problem 1

An excavated ancient Chinese bronze carillon was covered by rust containing

\ce{CuCl}

\ce{Cu2O}

and

\ce{Cu2(OH)3Cl}

. Simulation showed

\ce{CuCl}

forms first under air and Cl-containing water;

\ce{Cu2(OH)3Cl}

is then produced by routes (a)

\ce{CuCl -> Cu2O -> Cu2(OH)3Cl}

and (b,c)

\ce{CuCl -> Cu2(OH)3Cl}

directly. Standard Gibbs energies of formation (kJ mol⁻¹, 298 K):

\ce{Cu2O(s)}

−146,

\ce{CuO(s)}

−130,

\ce{CuCl(s)}

−120,

\ce{Cu2(OH)3Cl(s)}

−1338,

\ce{Cl^-(aq)}

−131,

\ce{OH^-(aq)}

−157,

\ce{H2O(l)}

−237. (1.1) (i) Write balanced equations for (a)

\ce{CuCl -> Cu2O}

, (b)

\ce{Cu2O -> Cu2(OH)3Cl}

and (c)

\ce{CuCl -> Cu2(OH)3Cl}

; (ii) calculate

\Delta_r G^0(298\ \mathrm{K})

for each; (iii) decide the direction of (a) in air at

c(\ce{HCl}) = 1.0\times10^{-4}

mol dm⁻³. (1.2) Rate constants of (c):

k_c = 1.29\times10^{-4}

mol dm⁻³ s⁻¹ at 25 °C and

2.50\times10^{-4}

at 40 °C: (i) find the activation energy; (ii) assign the reaction order; (iii) given that the rate-determining step is monolayer adsorption of

\ce{O2}

on CuCl, write the rate equation and the condition for the order in (ii). (1.3) A Cu plate is split; part Cu(1) is hammered. For the cell

\ce{Cu(1)}|\ce{CuSO4(aq)}|\ce{Cu(2)}

with

E = \Phi_R - \Phi_L

, is

E < 0

=0

>0

? Give the net cell reaction. (1.4) In a Cu–Zn alloy (

x_{\ce{Cu}} = 0.750

) with fcc structure and density 8.51 g cm⁻³, calculate the radius of the statistical atoms. (

A_r(\ce{Cu}) = 63.5

A_r(\ce{Zn}) = 65.4

Step 3 of 5: Direction of (a) in dilute HCl

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Intuition

Removing two product ions (H⁺ and Cl⁻) at 10⁻⁴ M each shifts the equilibrium by ~91 kJ mol⁻¹ — enough to overturn a +69 kJ mol⁻¹ standard barrier.

\Delta_r G_a = \Delta_r G^0_a + RT\ln[\ce{H+}]^2[\ce{Cl^-}]^2 = 69 + 2RT\ln(10^{-4})^2 = -22.3\ \mathrm{kJ\,mol^{-1}} < 0

Analysis

Although $\Delta_r G^0_a > 0$ , the product $[\ce{H+}]^2[\ce{Cl^-}]^2 = (10^{-4})^4 = 10^{-16}$ makes $Q$ tiny: $\Delta_r G_a = 69 + 2(8.314)(298)\ln 10^{-16}/1000 \approx -22.3$ kJ mol⁻¹, so (a) still runs to the right in dilute HCl.