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fgiga [73]
3 years ago
8

What is the final, balanced equation that is formed by combining these two half reactions?

Chemistry
1 answer:
lara31 [8.8K]3 years ago
7 0

Answer: Balanced equation that is formed by combining these two half reactions is Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O.

Explanation:

A chemical equation which contains same number of atoms on both reactant and product side is called a balanced chemical equation.

For example, Cu \rightarrow Cu^{2+} + 2e^{-}

NO^{-}_{3} + 2e^{-} + 2H^{+} \rightarrow NO^{-}_{2} + H_{2}O

On cancelling the common species from both these half-reactions, the complete balanced equation will be as follows.

Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O

Thus, we can conclude that balanced equation that is formed by combining these two half reactions is Cu + NO^{-}_{3} + 2H^{+} \rightarrow Cu^{2+} + NO^{-}_{2} + H_{2}O.

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Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) → 2SO3(g) Suppose the volume of this system is c
oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be  increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen  (i.e 2 moles +1 mole  =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be P_1

and the total pressure at the final equilibrium be P_2

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝  1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

P_1V_1 = P_2V_2

From the foregoing; since the volume is decreased to one- half to initial Volume; then ,

V_2 =  \dfrac{V_1}{\dfrac{3}{2}} ----- (1)

also;

Thus ;

P_1V_1 = P_2(  \dfrac{V_1}{\frac{3}{2}})

P_1V_1 = P_2 * 2  \dfrac{V_1}{3}

3 P_1 V_1 = 2 P_2 V_1

Dividing both sides by V_1

3P_1 = 2P_2

P_2 =P_1 \dfrac{3}{2}  ----- (2)

From ;

P_1V_1 = P_2V_2

P_2 V_2 = P_1 * \dfrac{3}{2}* \dfrac{V_1}{\frac{3}{2}}

P_2 V_2 = P_1 * \dfrac{3}{2}*   \dfrac{2 }{3}}*V_1

P_2 V_2 = P_1 V_1

Thus; The new equilibrium total pressure will be  increased to one-half to initial total pressure.

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If you had excess aluminum, how many moles of aluminum chloride could be produced from 28.0 g of chlorine gas, Cl2?
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