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katovenus [111]
4 years ago
14

1. 3A + 2B + C + 2D + 20 kJ

Chemistry
1 answer:
erma4kov [3.2K]4 years ago
3 0

Answer:

a) the reverse reaction is favoured

b) the forward reaction is favoured

c) the forward reaction is favoured

Explanation:

The equation ought to have been correctly written as;

3A + 2B --------> C + 2D. ∆H =20 kJ

Actually, we can see that the reaction is endothermic since ∆H= positive.

We know that when pressure is decreased, the reaction tends towards the side with higher total volumes. There are five volumes(moles) of reactants and three volumes(moles) of products. A decrease in pressure will favour the reverse reaction.

Being an endothermic reaction, increase in temperature is known to favour the forward reaction. Similarly, removing D will drive the equilibrium forward thereby favouring the forward reaction.

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vlada-n [284]
The given condition is STP, under this condition, gas has a rule of 22.4 L per mole. And the given equation is already balanced. The ratio of mole number is the same as the ratio of the volume and is also the same as the ratio of coefficients. So the answer is 4.0 liters. 
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3 years ago
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adoni [48]
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4 0
3 years ago
Does anyone know the answer to this? is it A?
AfilCa [17]

Answer:

1/2     ( the two answers given are not FRACTIONS)

Explanation:

Moles of  CH3OH   = 128 gm / 32 gm / mole   = 4 moles

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3 0
2 years ago
The activation energy of an uncatalyzed reaction is 95 kJ / mol. The addition of a catalyst lowers the activation energy to 55 k
dybincka [34]

Answer:

The rate of the catalyzed reaction will increase by a 1.8 × 10⁵ factor.

Explanation:

The rate of a reaction (r) is proportional to the rate constant (k). We can find the rate constant using the Arrhenius equation.

k=A.e^{-Ea/R.T}

where,

A: collision factor

Ea: activation energy

R: ideal gas constant

T: absolute temperature (125°C + 273 = 398 K)

For the uncatalized reaction,

kU=A.e^{-95\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=3.4\times 10^{-13}A

For the catalized reaction,

kC=A.e^{-55\times 10^{3} kJ/mol /(8.314J/K.mol).398K}=6.0\times 10^{-8}A

The ratio kC to kU is 6.0 × 10⁻⁸A/3.4 × 10⁻¹³A = 1.8 × 10⁵

8 0
4 years ago
2. The combined mass of a class of 25 students is about<br> 800 mg<br> 800 g<br> 800 kg<br> 8 t
Volgvan

Answer: i think that it was 800 mg

Explanation:

3 0
3 years ago
Read 2 more answers
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