The number of equations corresponds to the heat of formation at 298 K H -2 g -Br-21- rightarrow 2 HBr - g C graphite-s - O -2 g rightarrow CO -2 g C diamond -frac12 O -2 g - rightarrow -CO g

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Standard XII

Chemistry

Types of Equillibrium

The number of...

Question

The number of equations corresponds to the heat of formation at 298K
$H_{2 (g)} + B r_{2 (l)} \to 2 H B r_{(g)} C_{(d i a m o n d)} + 2 H_{2 (g)} \to C H_{4 (g)} C_{(g r a p h i t e, s)} + 2 H_{2} (g) + \frac{1}{2} O_{2} (l) \to C H_{3} O H (l) H_{2 (g)} + \frac{1}{2} O_{2 (g)} \to H_{2} O_{(l)} C_{(g r a p h i t e, s)} + 3 H_{2} (g) + \frac{1}{2} O_{2} (l) \to C_{2} H_{5} O H (l) \frac{1}{2} N_{2 (g)} + \frac{3}{2} H_{2 (g)} \to N H_{3 (g)} C_{(g r a p h i t e . s)} + O_{2 (g)} \to C O_{2 (g)} C_{(d i a m o n d)} + \frac{1}{2} O_{2 (g)} \to C O (g)$ ___

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Solution

Heat of formation : Enthalpy change when 1 mole of compound is formed from its most stable state of aggregation (also known as reference state) is called heat of formation.

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Similar questions

Q. The enthalpy of formation of reaction :

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) i s Δ_{r} H^{o} = - 572 k J m o l^{- 1}

What will be the enthalpy of formation of

H_{2} O (l)

Q. Consider the reactions

(i) C O (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g); K_{1} (i i) C H_{4} (g) + H_{2} O (g) ⇌ C O (g) + 3 H_{2} (g); K_{2} (i i i) C H_{4} (g) + 2 H_{2} O (g) ⇌ C O_{2} (g) + 4 H_{2} (g); K_{3}

which of the following is correct?

C l_{2} (g) + O H^{-} (a q) \to O C l^{-} (a q) + C l^{-} (a q)

Balance the above reaction using oxidation number method.

Question 14

Which of the following reactions is an example f use of water gaas in the synthesis of other compounds?

$(a) C H_{4} (g) + H_{2} O (g) 1270 K - --- \to N i C O (g) + H_{2} (g) (b) C O (g) + H_{2} O (g) 673 K - ---- \to C a t a l y s t C O_{2} (g) + H_{2} (g) (c) C_{n} H_{2 n + 2} + n H_{2} O (g) 1270 k - -- \to N i n C O + (2 n + 1) H_{2} (d) C O (g) + 2 H_{2} (g) C o b a l t - ---- \to C a t a l y s t C H_{3} O H (l)$

Q. Assertion :Maximum possible height attained by a projectile is

\frac{u^{2}}{2 g}

, where u is initial velocity Reason: To attain maximum height, a particle is thrown vertically upwards at

θ = 90^{o}

to the ground.

H = \frac{u^{2}}{2 g} s i n^{2} θ \Rightarrow H_{m a x} = \frac{u^{2}}{2 g}

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