Enthalpy of CH4 + $${1 \over 2}$$ O2 $$ \to $$ CH3OH is negative. If enthalpy of combustion of CH4 and CH3OH are x and y respectively. Then which relation is correct?
A
x > y
B
x < y
C
x = y
D
x3 y
Explanation
CH4 + $${1 \over 2}$$ O2 $$ \to $$ CH3OH
Given that for this reaction, $$\Delta $$Hr = -ve
CH4 + 2O2 → CO2 + 2H2O, $$\Delta $$H = x ....(1)
CH3OH + $${3 \over 2}$$
O2 → CO2 + 2H2O, $$\Delta $$H = y ....(2)
Most probable oxidation state of Pb and Sn will be
A
Pb4+, Sn4+
B
Pb4+, Sn2+
C
Pb2+, Sn2+
D
Pb2+, Sn4+
Explanation
PbO2 → PbO
Pb has +4 oxidation state In PbO2.
Pb has +2 oxidation state In PbO.
Here $$\Delta $$G is negative that is why reaction is spontaneous and
Pb4+ reduces Pb2+ thus, Pb2+ is most probable
oxidation state of Pb.
SnO2 $$ \to $$ SnO;
Sn has +4 oxidation state In SnO2.
Sn has +2 oxidation state In SnO.
$$\Delta $$G is positive that is why reaction is not spontaneous
thus, +4 oxidation state of Sn is more stable as it does
not change to +2 oxidation state spontaneously.
3
AIPMT 2001
MCQ (Single Correct Answer)
When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?
A
q = w = 500 J, $$\Delta $$E = 0
B
q = $$\Delta $$E = 500 J, w = 0
C
q = w = 500 J, $$\Delta $$E = 0
D
$$\Delta $$E = 0, q = w = $$-$$ 500 J
Explanation
As, $$\Delta $$E = q + W
Also, W = – p$$\Delta $$V
As $$\Delta $$V = 0
So, W = 0
$$ \Rightarrow $$ $$\Delta $$E = q
Now, q = 500 J
Thus, $$\Delta $$E = q = 500 J
4
AIPMT 2001
MCQ (Single Correct Answer)
Change in enthalpy for reaction,
2H2O2(l) $$ \to $$ 2H2O(l) + O2(g)
if heat of formation of H2O2(l) and H2O(l) are $$-$$188 and - 286 kJ/mol respectively, is