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2 years ago

CO + 2H2 CH3OH How many molecules of CO are needed to form 600g of CH3OH?

Chemistry

1 answer:

MA_775_DIABLO [31]2 years ago

8 0

Explanation:

molecules of Hydrogen gas to produce 1 molecule of ... 1 mol CO. 1 mol CO. 1 mol CH3OH. 2 mol H2. 1 mol CO. 1 mol CO. 1 mol CH3OH ... How many moles of hydrogen.

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Nitrogen monoxide, NO, reacts with hydrogen, H₂, according to the following equation.

Anvisha [2.4K]

Answer : The rate law for the overall reaction is, $Rate=k[NO]^2[H_2]$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

As we are given the mechanism for the reaction :

Step 1 : $2NO+H_2\rightarrow N_2+H_2O_2$ (slow)

Step 2 : $H_2O_2+H_2\rightarrow 2H_2O$ (fast)

Overall reaction : $2NO+2H_2\rightarrow N_2+2H_2O$

The rate law expression for overall reaction should be in terms of $NO\text{ and }H_2$ .

As we know that the slow step is the rate determining step. So,

The slow step reaction is,

$2NO+H_2\rightarrow N_2+H_2O_2$

The expression of rate law for this reaction will be,

$Rate=k[NO]^2[H_2]$

Hence, the rate law for the overall reaction is, $Rate=k[NO]^2[H_2]$

8 0

3 years ago

Which statement about chemical reaction rates is true?

aniked [119]

Answer:

Explanation:

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3 0

3 years ago

Read 2 more answers

In the reaction A + B C, doubling the concentration of A doubles the reaction rate and doubling the concentration of B does not

frutty [35]

Answer:

rate = k[A]

Explanation:

The equation that relate reaction rate with reactant concentrations is known as the rate law.

for a reaction:

A + B → C

the rate law can be expressed as:

Rate = k[A]ᵃ[B]ᵇ

The proportionality constant, k, is known as the rate constant, the powers a and b is the reaction order with respect to reactants A and B, respectively.

for this reaction doubling the concentration of A doubles the reaction rate that means

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

Rate₁ = k[A]₁ᵃ[B]ᵇ → eq. 1

Rate₂ = k[A]₂ᵃ[B]ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]₂ᵃ[B]ᵇ) / (k[A]₁ᵃ[B]ᵇ)

using

Rate₂ = 2 *Rate₁ and [A]₂ = 2 [A]₁

∴ (2 Rate₁ / Rate₁) = ( k [2]ᵃ[B]ᵇ) / (k[1]ᵃ[B]ᵇ)

(2) = (2)ᵃ
taking log of both sides
log (2) = a Log (2)
0.693 = a * 0.693
a =1

∴ order of reaction with respect to A is first (=1) → (1)

Doubling the concentration of B does not affect the reaction rate.

that means

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

Rate₁ = k[A]ᵃ[B]₁ᵇ → eq. 1

Rate₂ = k[A]ᵃ[B]₂ᵇ → eq. 2

Dividing eq. 2 by eq. 1 one can get

(Rate₂ / Rate₁) = (k [A]ᵃ[B]₂ᵇ) / (k[A]ᵃ[B]₁ᵇ)

using

Rate₂ = Rate₁ and [B]₂ = 2 [B]₁

∴ (Rate₁ / Rate₁) = ( k [A]ᵃ[2]ᵇ) / (k[A]ᵃ[1]ᵇ)

(1) = (2)ᵇ
taking log of both sides
log (1) = b Log (2)
0 = 0.693 * b
b = 0

∴ order of reaction with respect to B is zero → (2)

So, from 1 and 2 the right choice is rate = k[A]¹[B]⁰= k[A]

6 0

3 years ago

What is the mass percent of oxygen (O) in SO2?

REY [17]

Answer:

49.92%

Explanation:

4 0

3 years ago

At a constant temperature, a sample of helium at 760 torr in a closed container was compressed from 5.0 L to 3.0 L. What was the

borishaifa [10]

Answer:

The new pressure exerted by the He, is 1266.6 Torr

Explanation:

A typical problem of gases where the volume is increased and the moles and T° keeps on constant. This is an indirect proportion because when the volume of the flask is increased, pressue decreases because molecules collide to a lesser extent with the walls of the vessel.

P₁ . V₁ = P₂ . V₂

760 Torr . 5L = P₂ . 3L

P₂ = (760 Torr . 5L) / 3L

P₂ = 1266.6 Torr

7 0

3 years ago