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3 years ago

Calcutale Grxn for the following equation at 25°C: 4KClO3(s) → 3KClO4(s) KCl(s)

Chemistry

1 answer:

Step2247 [10]3 years ago

5 0

Answer:

-133.2 kJ

Explanation:

Let's consider the following balanced equation.

4 KClO₃(s) → 3 KClO₄(s) + KCl(s)

We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression.

ΔG°rxn = 3 mol × ΔG°f(KClO₄(s)) + 1 mol × ΔG°f(KCl(s)) - 4 mol × ΔG°f(KClO₃(s))

ΔG°rxn = 3 mol × (-303.1 kJ/mol) + 1 mol × (-409.1 kJ/mol) - 4 mol × (-296.3 kJ/mol)

ΔG°rxn = -133.2 kJ

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3 years ago

200 g of solid sodium hydroxide (NaOH) are dissolved in 750 mL of water. What is the concentration of this solution?

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Answer:

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Explanation:

Molarity = $\frac{mol}{L}$

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Read 2 more answers

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are remo

frozen [14]

Answer:

The total work is 4957.45J

Explanation:

For an ideal gas, at constant temperature the definition of work (W) is

$W = - P.dV = - n.R.T \int\limits^i_f {\frac{dV}{V} \\W = - n.R.T. Ln (\frac{V_{f}}{V_{I}})\\W = - n.R.T. Ln (\frac{P_{i}}{P_{f}})$

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

$W_{1} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{5.50atm}{2.43atm}) = 23.44atm.Lx101.325\frac{J}{atm.L} =2375.44J$

Stape 2: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

$W_{2} = - 1molx0.082\frac{atm.L}{K.mol}x350KxLn (\frac{2.43atm}{1.00atm}) = 25.48atm.Lx101.325\frac{J}{atm.L} =2582.01J$

The total work is the sum of the two steps

$W = W_{1} + W_{2} = 2375.44J + 2582.01J = 4957.45J$

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3 years ago