Which is correct order for bond energy?

N_{2} > H F > O_{2} > H_{2}

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N_{2} > O_{2} > H F > H_{2}

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N_{2} > H_{2} > O_{2} > H F

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H F > N_{2} > O_{2} > H_{2}

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Solution

The correct option is A $N_{2} > H F > O_{2} > H_{2}$
The correct order for bond energy is $N_{2} > H F > O_{2} > H_{2}$ .
Nitrogen molecule contains triple bond thus have highest bond energy followed by oxygen which contains double bond , then hydrogen which contains single bond.But bond energy of $H F$ which contains single bond is higher than $O_{2}$ due to large difference of electronegativity between hydrogen and fluorine atom.
B.E; $N_{2} > H F > O_{2} > H_{2}$
kcal/mol $225.08 > 134.6 > 118.32 > 1.4.18$

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Q. At

S T P

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H_{2}, N_{2}, O_{2}

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Q. Gases

H_{2} H e O_{2} N_{2}

Critical temperature(K) 33.2 5.3 154.3 126

From the above data what whould be the order of liquefaction of these gases?
Start writing the order from the gas liquefying first.

Gases posses charcteristic critical temperature which depends upo the magnitude of intermolecular forces between the particles. Following are the critical temperatures of some gases.

$\begin{matrix} G a s e s & H_{2} & H e & O_{2} & N_{2} C r i t i c a l t e m p e r a t u r e i n K e l v i n & 33.2 & 5.3 & 154.3 & 126 \end{matrix}$

From the above data what would be the order of liquefaction of these gases? Start writing the order from the gas liquefying first

(a) $H_{2}, H e, O_{2}, N_{2}$ (b) $H e, O_{2}, H_{2}, N_{2}$

The possible mechanism for the reaction:

$2 H_{2 +} 2 N O \to N_{2} + 2 H_{2} O$ is

$(i) 2 N O ⇌ N_{2} O_{2} (i i) N_{2} O_{2} + H_{2} \to N_{2} O + H_{2} O (slow) (i i i) N_{2} O + H_{2} \to N_{2} H_{2} O (fast)$ What is (i) the rate law for the reaction (ii) the order of the reaction?

Balance the following equations :

(a) $N_{2} + H_{2} \to N H_{3}$

(b) $H_{2} + O_{2} \to H_{2} O$

(d) $C O + O_{2} \to C O_{2}$

(e) $Z n + H C I \to Z n C l_{2} + H_{2}$

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