Question

number two a and b

2. a) Fill in the table below to indicate the sign (positive or negative or zero or 0, unable to determine or unknown) of the indicated value for each of the systems described below. Assume that systems not involving gases have negligible expansion or compression System #: A reaction occurs under standard conditions, and the vessel containing it becomes Warm during the reaction. Initial state reactants Final state products in thermodynamic equilibrium Susten 2: In proteins, Zn is often observed as a square planar complex, where the four ligands are the atoms of cysteine amino acids. The relevant reaction is Zn(H20), "(aq) + 4 cysteine(aq) → Zn(cysteine)/(aq) + 6 H 0(1) A for Zn-S is 205 kJ mol'' and How for Zn-Ois-160 kJ mol". The reaction occurs spontaneously under physiological conditions (which are not standard). Initial state reactants. Final state: products, System ANTAU NYASAGTAG b) For each system, brielly explain why you indicated such signs for each thermodynamic variable

Answer 1

1 a. ) The answer is fill in the boxes and explanation is given in

System	∆V	∆U	∆H	∆S	∆G	∆Go
#1	0	-	-	-	0	-
#2	0	0	-	+	-	0

Reason:

For system 1

∆V

The system reaction occuring in liquid or solid phase , so there volume is negligible change hence ∆V = 0

∆U

The internal enrgy depend on temperature or ∆qbecause

∆U = -p∆V + ∆q

here ∆V is zero so first term is zero ,

now ∆U = ∆q

the vessel is warm so ∆q is negative

henve ∆U = negative

For ∆H

at constant pressure heat is called ∆H

so reaction is exothermic hence ∆H = negative

For ∆S = ∆q/T= negative value of ∆q divide by temperature= negative

For ∆G

we know that at equilibrium no change in gibbs energy

∆G = 0

For ∆G^o

we know that

∆G^o = ∆H^o -T∆S^o= so negative value of ∆H^o- T negative value of ∆S^o

so overall ∆G^o = negative because sytem have hetaing so ∆H^o is more changing than∆S^o so overall negative.

For syatem 2

∆V

The system reaction occuring in liquid or solid phase , so there volume is negligible change hence ∆V = 0

∆U

The internal enrgy depend on temperature or ∆qbecause

∆U = -p∆V + ∆q

here ∆V is zero so first term is zero ,

now ∆U = ∆q

the vessel temperature note change so ∆q = 0

henve ∆U = 0

For ∆H

∆H=H(products)-H(reactant) = 4*(-205kJmol^-1)- 6*(-160kJmol^-1) = (-1000+960)kJmol^-1) = -40kJmol^-1)

hence ∆H = negative

For ∆S =in final porduct seven molecule and in reactant five molecule = positve

For ∆G

we know that

∆G = ∆H -T∆S= so negative value of ∆H- T positive value of ∆S

so overall ∆G = negative

For ∆G^o

we know that

∆G^o = -RTlnK

here equilibrium is not occur so we consider K= 0

so ∆G^o = negative