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laila [671]
2 years ago
11

When a reaction has reached equilibrium... A. Q = K B. The rate of the forward reaction and the rate of the reverse reaction are

equal. C. Both the forward and reverse reactions stop occurring. D. The concentration of products and the concentration of reactants are equal. E. The forward rate constant and the reverse rate constant are equal.
Chemistry
1 answer:
grin007 [14]2 years ago
4 0

Answer:

The correct answer is option A.

Explanation:

Equilibrium is a state when rate of forward reaction is equal to the rate of backward reaction. The concentration of reactants and products becomes constant at this state.

The ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric coefficients is termed as Equilibrium constant. It is denoted by K_{eq}.

aA + bB \rightleftharpoons cC

K_{eq}=\frac{[C]^c}{[A]^a[B]^b}

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The volume of chlorine molecules produced at STP would be 96 dm³.

<h3>Stoichiometric problem</h3>

Sodium chloride ionizes during electrolysis to produce sodium and chlorine ions as follows:

NaCl --- > Na^+ + Cl^-

This means that 1 mole of sodium chloride will produce 1 mole of sodium ion and 1 mole of chlorine ion respectively.

Recall that: mole = mass/molar mass

Hence, 234 g of sodium chloride will give:

                         234/58.44 = 4.00 moles.

Thus, the equivalent number of moles of chlorine produced by 234 g of sodium chloride will be 4 moles.

Recall that:

1 mole of every gas at Standard Temperature and Pressure = 24 Liters.

Hence:

4 moles of chlorine = 4 x 24 = 96 Liters or 96 dm³.

More on stoichiometric problems can be found here: brainly.com/question/14465605

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A sample of nitrogen is initially at a pressure of 1.7 kPa, a temperature of -10 C and a volume of 7.5 m3. Then the volume is de
zhannawk [14.2K]

Answer:

\boxed{\text{2.6 kPa}}

Explanation:

To solve this problem, we can use the Combined Gas Laws:

\dfrac{p_{1}V_{1} }{T_{1}} = \dfrac{p_{2}V_{2} }{T_{2}}

Data:

p₁ = 1.7 kPa; V₁ = 7.5 m³;  T₁ =   -10 °C

p₂ = ?;          V₂ = 3.8 m³; T₂ = 200  K

Calculations:

(a) Convert temperature to kelvins

T₁ = (-10   + 273.15) K = 263.15 K

(b) Calculate the pressure

\begin{array}{rcl}\dfrac{1.7 \times 7.5 }{263.15} & = & \dfrac{p_{2} \times 3.8}{200}\\\\0.0485 & = & 0.0190p_{2}\\p_{2} & = & \textbf{2.6 kPa}\\\end{array}\\\text{The new pressure of the gas is \boxed{\textbf{2.6 kPa}}}

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