Question

A reaction gives the following data:

K                                                                     T

3.52 x 10^-7                                                10C

3.02 x 10^-5                                                83C

2.19 x 10^-4                                                120C

1.16 x 10^-3                                                154C

3.95 x 10^-2                                                235C

A.) what is the reagent used to separate all 5 ions along with the reagents they separate? Explain.

B) Solve for Delta H and Delta S Graphically. Excel program or any other computer generated graph will not be accepted.

C) Solve for Delta H using the 2 point form

** Please explain with as much detail as possible , where applicable**

Answer 1

B) Using,

ln(k2/k1) = dH/R[1/T1 - 1/T2]

with,

data from T1 = 120 oC and T2 = 154 oC

we get,

ln(1.16 x 10^-3/2.19 x 10^-4) = dH/8.314[1/(273 + 120) - 1/(273 + 154)]

dH = 68.409 kJ/mol

Now,

ln(k) = -dH/R(1/T) + dS/R

ln(2.19 x 10^-4) = -68409/8.314 x 393 + dS/8.314

dS = 103.98 J/mol.K

C) dH by T1 = 120 oC and T2 = 154 oC

Using,

ln(k2/k1) = dH/R[1/T1 - 1/T2]

with,

data from T1 = 120 oC and T2 = 154 oC

we get,

ln(1.16 x 10^-3/2.19 x 10^-4) = dH/8.314[1/(273 + 120) - 1/(273 + 154)]

dH = 68.409 kJ/mol

Alternatively, we may plot 1/T (K-1) on x-axis and ln(k) on y-axis. We get a straight line. Slope = -dH/R