### What Is The Rate Law For The Following Mechanism In Terms Of The Overall Rate Constant K?

- Marvin Harvey
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The rate law is rate=k2.

## What is the rate law constant k?

Summary –

- A rate law is an expression showing the relationship of the reaction rate to the concentrations of each reactant.
- The specific rate constant \(\left( k \right)\) is the proportionality constant relating the rate of the reaction to the concentrations of reactants.
- The rate law and the specific rate constant for any chemical reaction must be determined experimentally.

#### What is the rate law for the overall reaction that is consistent with the proposed mechanism?

The rate law that is consistent with the mechanism is rate = k2. The rate law for the reaction represented by the equation above is rate = k.

### How do you find the rate law constant?

An example of how to find the rate constant equation, where r=k2 r = k 2, for instance, is to solve for k, which gives k=r2 k = r 2 as the rate constant equation.

## What is the rate constant at 760 k?

The rate constant of a reaction at 700 K and 760 K are 0.011 M^-1 s^-1, respectively.

### How do you find the K value in chemistry?

Combining Equilibrium Constant Expressions – Chemists frequently need to know the equilibrium constant for a reaction that has not been previously studied. In such cases, the desired reaction can often be written as the sum of other reactions for which the equilibrium constants are known.

The equilibrium constant for the unknown reaction can then be calculated from the tabulated values for the other reactions. To illustrate this procedure, let’s consider the reaction of \(N_2\) with \(O_2\) to give \(NO_2\). This reaction is an important source of the \(NO_2\) that gives urban smog its typical brown color.

The reaction normally occurs in two distinct steps. In the first reaction (1), \(N_2\) reacts with \(O_2\) at the high temperatures inside an internal combustion engine to give \(NO\). The released \(NO\) then reacts with additional \(O_2\) to give \(NO_2\) (2).

- \(N_ +O_ \rightleftharpoons 2NO_ \;\; K_1=2.0 \times 10^ \)
- \(2NO_ +O_ \rightleftharpoons 2NO_ \;\;\;K_2=6.4 \times 10^9\)

Summing reactions (1) and (2) gives the overall reaction of \(N_2\) with \(O_2\):

\(N_ +2O_ \rightleftharpoons 2NO_ \;\;\;K_3=?\)

The equilibrium constant expressions for the reactions are as follows: \ What is the relationship between \(K_1\), \(K_2\), and \(K_3\), all at 100°C? The expression for \(K_1\) has \(^2\) in the numerator, the expression for \(K_2\) has \(^2\) in the denominator, and \(^2\) does not appear in the expression for \(K_3\).

Multiplying \(K_1\) by \(K_2\) and canceling the \(^2\) terms, \}=\dfrac =K_3 \label \] Thus the product of the equilibrium constant expressions for \(K_1\) and \(K_2\) is the same as the equilibrium constant expression for \(K_3\): \ The equilibrium constant for a reaction that is the sum of two or more reactions is equal to the product of the equilibrium constants for the individual reactions.

In contrast, recall that according to Hess’s Law, \(ΔH\) for the sum of two or more reactions is the sum of the ΔH values for the individual reactions. To determine \(K\) for a reaction that is the sum of two or more reactions, add the reactions but multiply the equilibrium constants.

- \(CO_ +3H_ \rightleftharpoons CH_ +H_2O_ \;\;\;K_1=9.17 \times 10^ \)
- \(CH_ +2H_2S_ \rightleftharpoons CS_ +4H_ )\;\;\; K_2=3.3 \times 10^4\)

Calculate the equilibrium constant for the following reaction at the same temperature.

\(CO_ +2H_2S_ \rightleftharpoons CS_ +H_2O_ +H_ \;\;\; K_3=?\)

Given : two balanced equilibrium equations, values of \(K\), and an equilibrium equation for the overall reaction Asked for : equilibrium constant for the overall reaction Strategy : Arrange the equations so that their sum produces the overall equation.

- If an equation had to be reversed, invert the value of \(K\) for that equation.
- Calculate \(K\) for the overall equation by multiplying the equilibrium constants for the individual equations.
- Solution : The key to solving this problem is to recognize that reaction 3 is the sum of reactions 1 and 2: \ \ \ The values for \(K_1\) and \(K_2\) are given, so it is straightforward to calculate \(K_3\): \ Exercise \(\PageIndex \) In the first of two steps in the industrial synthesis of sulfuric acid, elemental sulfur reacts with oxygen to produce sulfur dioxide.

In the second step, sulfur dioxide reacts with additional oxygen to form sulfur trioxide. The reaction for each step is shown, as is the value of the corresponding equilibrium constant at 25°C. Calculate the equilibrium constant for the overall reaction at this same temperature.

- \(\frac S_ +O_ \rightleftharpoons SO_ \;\;\; K_1=4.4 \times 10^ \)
- \(SO_ +\frac O_ \rightleftharpoons SO_ \;\;\; K_2=2.6 \times 10^ \)
- \(\frac S_ +\frac O_ \rightleftharpoons SO_ \;\;\; K_3=?\)

Answer : \(K_3 = 1.1 \times 10^ \)

### What is rate law explain with example?

Rate Law In a chemical reaction, it is important to consider not only the chemical properties of the reactants, but also the conditions under which the reaction occurs, the mechanism through which it takes place, the rate at which it occurs, and the equilibrium toward which it is proceeding.

- The rate law of a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentrations of the reactants participating in it.
- Example:
- If a reaction is given by $aA + bB \to cC + dD$
- Where a, b, c, and d denotes the stoichiometric coefficients of the reactants and products, the rate equation for the reaction is given by:
- $Rate \propto ^x^y$
- $\Rightarrow Rate = k^x^y$

- $$ & $$ denote the concentrations of the reactant side $A$ & $B$.
- The proportionality constant $‘k’$ is the rate constant for the reaction.
- $x$ & $y$ denote the partial reaction orders for the reactant side $A$ & $B$ (This may or may not be equal to their stoichiometric coefficients $a$ & $b$).

Note: The expression of the rate law for a specific reaction can only be determined experimentally. The rate law expression is not determined by the balanced chemical equation.

#### What is the rate law expression for the overall reaction if step 1 is the slow step?

Solution – A The rate law for step 1 is rate = k 1 2 ; for step 2, it is rate = k 2, B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k 1 2, This is the same as the experimentally determined rate law.

#### What is the rate law for 2NO 2H2 N2 2H2O?

The rate law for the reaction 2H 2(g) + 2NO (g) → N 2(g) + 2H 2 O (g) is given by rate = k 2, The reaction occurs in the following two steps : (i) H 2(g) + 2NO (g) → N 2 O (g) + H 2 O (g) (ii) N 2 O (g) + H 2(g) → N 2(g) + H 2 O (g) What is the role of N 2 O in the mechanism? What is the molecularity of each of the elementary steps?

#### What is rate constant example?

The rate constant is specific for a reaction under specific conditions, like temperature. Therefore, every chemical reaction has a variable rate constant. The reaction order is the summation of the partial reaction orders given for each reactant. For the example rate law equation, the reaction order = m + n + o.

## How is the rate constant k1 and k2 calculated?

Activation energy (Ea) and rate constant (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by Arrhenius equation which can be represented as either lnkk=−RE(T1−T1)

### How do you calculate K in Gibbs?

Temperature Dependence of the Equilibrium Constant – The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship is shown explicitly in Equation \(\ref \), which can be rearranged as follows: \ Assuming ΔH° and ΔS° are temperature independent, for an exothermic reaction (ΔH° 0), the magnitude of K increases with increasing temperature.

- The quantitative relationship expressed in Equation \(\ref \) agrees with the qualitative predictions made by applying Le Chatelier’s principle.
- Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K.

Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation \(\ref \) also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature.

In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence. If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K 1 and K 2 are the equilibrium constants for a reaction at temperatures T 1 and T 2, respectively.

Applying Equation \(\ref \) gives the following relationship at each temperature: \ Subtracting \(\ln K_1\) from \(\ln K_2\), \ Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K 1 ) allow us to calculate the value of the equilibrium constant at any other temperature (K 2 ), assuming that ΔH° and ΔS° are independent of temperature.

- If ΔG° 1, and products are favored over reactants at equilibrium.
- If ΔG° > 0, then K < 1, and reactants are favored over products at equilibrium.
- If ΔG° = 0, then K = 1, and the amount of products will be roughly equal to the amount of reactants at equilibrium. This is a rare occurrence for chemical reactions.

If a system is not at equilibrium, ΔG and Q can be used to tell us in which direction the reaction must proceed to reach equilibrium. ΔG is related to Q by the equation \(ΔG= RT\ln \dfrac \).

- If ΔG Q, and the reaction must proceed to the right to reach equilibrium.
- If ΔG > 0, then K < Q, and the reaction must proceed to the left to reach equilibrium.
- If ΔG = 0, then K = Q, and the reaction is at equilibrium.

We can use the measured equilibrium constant K at one temperature, along with ΔH° to estimate the equilibrium constant for a reaction at any other temperature.

## What will the rate constant be for this reaction at 700 K?

The rate constants of a reaction at 500 K and 700 K are `0.02 s^(–1) and 0.07 s^(–1)`, respectively.

### What is k in chemistry?

K is the rate constant that determines the relation between the molar concentration of reactant and the rate of reaction. It depends on the temperature. For a reaction that is at equilibrium we use K.

#### What is k in law of mass action?

Recommended Videos –

- For a balanced reaction of the type,
- aA + bB ⇌ cC + dD
- According to the law of mass action, the constant value obtained by relating equilibrium concentrations of reactants and products is called the, For the forward reaction, this is given by
- \(\begin K_c=\frac \end \)
- The equilibrium constant for the reverse reaction is the inverse of the forward reaction and is given by:
- \(\begin K’_c=\frac = \frac \end \)

If the coefficients of the chemical equation are multiplied by a factor ‘n’ then the equilibrium constant is raised by the power ‘n’ i.e. the constant becomes K c n,

Equilibrium Constant Representation | Expressed in terms of | Expressed as |

K c | Concentrations of reactants and products | \(\begin \frac \end \) |

K p | Partial pressures of reactants and products. (only for the substances which are in gaseous state) | \(\begin \frac ^ p_ ^ } ^ p_ ^ }\end \) |

K x | Mole fractions of reactants and products | \(\begin \frac \end \) |

#### How is the rate constant k1 and k2 calculated?

Activation energy (Ea) and rate constant (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by Arrhenius equation which can be represented as either lnkk=−RE(T1−T1)