For silver, CP( K-1 mol-1) = 23 + 0.01T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of H will be close to :
21 kJ
62 kJ
16 kJ
13 kJ
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero)
Isothermal process : q = -w
Cyclic process : q = -w
Isochoric process : = q
Adiabatic process : = -w
During compression of a spring the work done is 10 kJ and 2 kJ escaped to the surroundings as heat. The change in internal energy, U (in kJ) is:
-12
8
12
-8
Consider the given plot of enthalpy of the following reaction between A and B. A + B → C + D. Identify the incorrect statement.
C is the thermodynamically stable product.
Formation of A and B from C has highest enthalpy of activation.
Activation enthalpy to form C is 5 kJ mol-1 less than that to form D.
D is kinetically stable product.
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 J K–1 mol–1, calcuIate ΔU and ΔPV for this process. ( R= 8.0 JK-1 mol-1 )
ΔU = 14kJ; Δ(pV) = 18kJ
ΔU = 14kJ; Δ(pV) = 0.8J
ΔU = 14kJ; Δ(pV) = 4kJ
ΔU = 2.8kJ; Δ(pV) = o.8J
Among the following, the set of parameters that represents path functions, is:
(A) q+w
(B) q
(C) w
(D) H-TS
A, B and C
B, C and D
B and C
A and D
A process will be spontaneous at all temperatures if :
ΔH <0 and ΔS <0
ΔH <0 and ΔS >0
ΔH >0 and ΔS >0
ΔH >0 and ΔS <0
The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is:
Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s), E° = 2V at 298 K
-192
384
-384
192
The difference between ΔH and ΔU ( ΔH- ΔU), when the combustion of one mole heptane(I) is carried out a temperature T, is equal to:
-3RT
-4RT
3RT
4R
The correct thermodynamic conditions for the spontaneous reaction at all temperatures is
B.
ΔH< 0 and ΔS > 0According to the Gibbs-Helmholtz reaction for spontaneity as,
ΔG =ΔH-TΔS
For reaction to being spontaneous, ΔG must be negative.
For this ΔH should be negative ΔS should be positive.
therefore,
ΔHΔS >0.