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Increase of pressure favours the forward...

Increase of pressure favours the forward reaction in the following equilibrium

A

`H_(2(g)) +I_(2(g)) hArr 2HI_((g))`

B

`2NO_(2(g)) hArr N_(2)O_(4(g))`

C

`PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g))`

D

`N_(2(g)) + O_(2(g)) hArr 2NO(g)`

Text Solution

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The correct Answer is:
To determine which equilibrium reaction is favored by an increase in pressure, we need to analyze the number of moles of gas on both sides of the equilibrium. According to Le Chatelier's principle, if the pressure of a system at equilibrium is increased, the system will shift in the direction that reduces the number of gas molecules, thereby decreasing the pressure. ### Step-by-Step Solution: 1. **Identify the Reactions**: We have several reactions to analyze: - \( H_2 + I_2 \rightleftharpoons 2 HI \) - \( 2 NO_2 \rightleftharpoons N_2O_4 \) - \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \) - \( N_2O + O_2 \rightleftharpoons 2 NO \) 2. **Count Moles of Gas**: For each reaction, we will count the number of moles of gas on both sides: - For \( H_2 + I_2 \rightleftharpoons 2 HI \): - Reactants: 1 (H2) + 1 (I2) = 2 moles - Products: 2 (HI) = 2 moles - For \( 2 NO_2 \rightleftharpoons N_2O_4 \): - Reactants: 2 (NO2) = 2 moles - Products: 1 (N2O4) = 1 mole - For \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \): - Reactants: 1 (PCl5) = 1 mole - Products: 1 (PCl3) + 1 (Cl2) = 2 moles - For \( N_2O + O_2 \rightleftharpoons 2 NO \): - Reactants: 1 (N2O) + 1 (O2) = 2 moles - Products: 2 (NO) = 2 moles 3. **Determine the Effect of Pressure**: - According to Le Chatelier's principle, an increase in pressure will favor the side with fewer moles of gas: - For \( H_2 + I_2 \rightleftharpoons 2 HI \): No change (2 moles on both sides) - For \( 2 NO_2 \rightleftharpoons N_2O_4 \): Favors forward reaction (2 moles to 1 mole) - For \( PCl_5 \rightleftharpoons PCl_3 + Cl_2 \): Favors reverse reaction (1 mole to 2 moles) - For \( N_2O + O_2 \rightleftharpoons 2 NO \): No change (2 moles on both sides) 4. **Conclusion**: The only reaction that is favored by an increase in pressure is: \[ 2 NO_2 \rightleftharpoons N_2O_4 \] This is because it reduces the number of moles from 2 to 1, thus decreasing the pressure. ### Final Answer: The equilibrium reaction that is favored by an increase in pressure is: \[ 2 NO_2 \rightleftharpoons N_2O_4 \]
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