Consider the given equilibrium, $2 H_{2} O (l) + h e a t ⇌ 2 H_{2} (g) + O_{2} (g)$ , which of the following changes will decrease the equilibrium amount of water?

Adding more oxygen

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Increasing the temperature at constant pressure

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Increasing the pressure at constant temperature

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Adding catalyst

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Solution

The correct option is B Increasing the temperature at constant pressure
According to Le-Chatelier’s principle, for endothermic reactions increase in temperature shifts the equilibrium in forward direction. Hence, it will result in decrease in equilibrium amount of water.

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Similar questions

2 H_{2} O (l) ⇌ 2 H_{2} (g) + O_{2} (g)

Δ H = 241.7 k J

2 H_{2} O (l) + h e a t ⇌ 2 H_{2} (g) + O_{2} (g)

2 H_{2} O (l) ⇌ 2 H_{2} (g) + O_{2} (g)

Δ H = 241.7 k J

2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g)

H_{2} O ?

C_{(s)} + O_{2 (g)} \to C O_{2 (g)};

Δ H = - 394 k J

2 H_{2 (g)} + O_{2 (g)} \to 2 H_{2} O_{(l)};

Δ H = - 568 k J

C H_{4 (g)} + 2 O_{2 (g)} \to C O_{2 (g)} + 2 H_{2} O_{(l)};

Δ H = - 394 k J

C H_{4}

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