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kkurt [141]
2 years ago
9

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.

is there anything wrong with it

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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:

C

Explanation:

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3 0
3 years ago
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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
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