Ionic equation for Anode positive electrode

Chemistry

1 answer:

Karo-lina-s [1.5K]3 years ago

3 0

Answer:

A half-equation shows you what happens at one of the electrodes during electrolysis. Electrons are shown as e-. A half-equation is balanced by adding, or taking away, a number of electrons equal to the total number of charges on the ions in the equation.

Explanation:

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A sample of ammonia gas was allowed to come to equilibrium at 400 K. 2NH3(g) <---> N2(g) 3H2(g) At equilibrium, it was fou

Softa [21]

Answer:

Kc for this equilibrium is 2.30*10⁻⁶

Explanation:

Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products are held constant.

Being:

aA + bB ⇔ cC + dD

the equilibrium constant Kc is defined as:

$Kc=\frac{[C]^{c}*[D]^{d} }{[A]^{a} *[B]^{b} }$

In other words, the constant Kc is equal to the multiplication of the concentrations of the products raised to their stoichiometric coefficients by the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients. Kc is constant for a given temperature, that is to say that as the reaction temperature varies, its value varies.

In this case, being:

2 NH₃(g) ⇔ N₂(g) + 3 H₂(g)

the equilibrium constant Kc is:

$Kc=\frac{[N_{2} ]*[H_{2} ]^{3} }{[NH_{3} ]^{2} }$