What is the entropy change (in J K$$-$$1 mol$$-$$1) when one mole of ice is converted into water at 0oC? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol$$-$$1 at 0oC).
The molar heat capacity of water at constant pressure, C, is 75 J K$$-$$1 mol$$-$$1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is
A
1.2 K
B
2.4 K
C
4.8 K
D
6.6 K
Explanation
As we know, q = nC$$\Delta $$T
q = 1.0 kJ = 1000 J
C = 75 JK–1 mol–1
m = 100 g ⇒ Number of moles = $${{100} \over {18}}$$ g mol–1
Formation of solution from two components can be considered as
(i) Pure solvent $$ \to $$ separated solvent molecules, $$\Delta $$H1 (ii) Pure solute $$ \to $$ separated solute molecules, $$\Delta $$H2 (iii) Separated solvent and solute molecules $$ \to $$ solution, $$\Delta $$H3 Solution so formed will be ideal if
For which one of the following equations is $$\Delta $$Horeact equal to $$\Delta $$Hof for the product ?
A
Xe(g) + 2F2(g) $$ \to $$ XeF4(g)
B
2CO(g) + O2(g) $$ \to $$ 2CO2(g)
C
N2(g) + O3(g) $$ \to $$ N2O3(g)
D
CH4(g) + 2Cl2(g) $$ \to $$ CH2Cl2(l) + 2HCl(g)
Explanation
Heat of formation, $$\Delta $$Hof
of a substance is the
amount of heat absorbed or released when one
mole of this substance is formed directly from its
constituent elements.
In option (a), one mole of XeF4 is formed from
its constituent elements i.e., Xe and F2 thus, the
equation has equal value of $$\Delta $$Hor and $$\Delta $$Hof
.
In
option (b), the constituent atoms should be carbon
and oxygen only but the reactant used is CO thus,
$$\Delta $$Hor $$ \ne $$ $$\Delta $$Hof
In option (c), the reactant used is O3 which is again
not in its element form thus,
$$\Delta $$Hor $$ \ne $$ $$\Delta $$Hof
In option (d), two products are formed thus
$$\Delta $$Hor $$ \ne $$ $$\Delta $$Hof
Questions Asked from Thermodynamics
On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
