In the equilibrium reaction- N2 - 3H2- 2NH3- the sign of - H accompanying the reaction is-

The correct option is C negativeThe reaction for the formation of ammonia is N_2 + 3H_2⇔ nbsp;2NH_3+92.2 nbsp;kJ . Thus the reaction is exothermic an ...

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Standard XII

Chemistry

Difference between Internal Energy and Enthalpy

In the equili...

Question

In the equilibrium reaction, $N_{2} + 3 H_{2} \Leftrightarrow 2 N H_{3}$ , the sign of $Δ H$ accompanying the reaction is;

positive

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negative

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may be positive or negative

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cannot be predicted

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Solution

The correct option is C negative
The reaction for the formation of ammonia is $N_{2} + 3 H_{2} \Leftrightarrow 2 N H_{3} + 92.2 k J$ . Thus the reaction is exothermic and the sign of $Δ H$ is negative, i.e $Δ H = - 92.2 k J$ .

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In the equilibrium reaction, N2+3H2⇔2NH3, the sign of ΔH accompanying the reaction is;

The correct option is C negativeThe reaction for the formation of ammonia is N2+3H2⇔2NH3+92.2kJ. Thus the reaction is exothermic and the sign of ΔH is negative, i.e ΔH=−92.2kJ.

In the equilibrium reaction, $N_{2} + 3 H_{2} \Leftrightarrow 2 N H_{3}$ , the sign of $Δ H$ accompanying the reaction is;

The correct option is C negative
The reaction for the formation of ammonia is $N_{2} + 3 H_{2} \Leftrightarrow 2 N H_{3} + 92.2 k J$ . Thus the reaction is exothermic and the sign of $Δ H$ is negative, i.e $Δ H = - 92.2 k J$ .