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2 years ago

What controls the volumes of water in each cylinder once equilibrium is established?

1 answer:

6 0

The increase in volume shift the equilibrium towards making more moles of gas, decrease in volume shift the equilibrium towards producing fewer moles of gas.

<h3>What is equilibrium?</h3>

The equilibrium of the gas is based on the pressure of the gas. With the increase in pressure, the equilibrium moves towards making fewer molecules of gas.

The gas equilibrium is proportional to the equilibrium K.

Increase in volume shift the equilibrium towards making more moles of gas, decrease in volume shift the equilibrium towards producing fewer moles of gas.

Learn more about equilibrium

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Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:

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The above question is incomplete, here is the complete question:

Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o = 66.4 kJ$

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Answer:

The standard molar enthalpy of formation of NO is 90.25 kJ/mol.

Explanation:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ$

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To calculate the standard molar enthalpy of formation

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$ ...[3]

Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2] = [3]

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$

$\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2}$

$\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ$

According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :

= $\frac{\Delta H^o_{3}}{2 mol}$

$=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol$

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