CH₄ + $${1 \over 2}$$ O₂ $$ \to $$ CH₃OH

Given that for this reaction, $$\Delta $$H_r = -ve

CH₄ + 2O₂ → CO₂ + 2H₂O,    $$\Delta $$H = x ....(1)

CH₃OH + $${3 \over 2}$$ O₂ → CO₂ + 2H₂O,    $$\Delta $$H = y ....(2)

Eqn. (1) – Eqn. (2)

CH₄ + $${1 \over 2}$$ O₂ $$ \to $$ CH₃OH

$$\Delta $$H_r = x - y = -ve

$$ \therefore $$ x < y

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.

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

2H₂O_2(l) $$ \to $$ 2H₂O_(l) + O_2(g), $$\Delta $$H_r = ?

H_2(g) + O_2(g) → H₂O_2(l), $$\Delta $$H = – 188 kJ mol ...(1)

H_2(g) + O_2(g) → H₂O_(l), $$\Delta $$H = –286 kJ/mol ....(2)

(1) - (2)

H₂O_2(l) → H₂O_(l) + $${1 \over 2}$$O_2(g)

$$\Delta $$H = –286 – (– 188) = –98 kJ mol^–1

Multiplying this equation by 2 we get the required equation

2H₂O_2(l) → 2H²O_(l) + O_2(g)

$$\Delta $$H = 2 (– 98) kJ/mol = – 196 kJ/mol