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egoroff_w [7]
3 years ago
8

For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. The reaction is endothermic. What do yo

u expect to happen to the concentration of NO if the temperature is doubled? 2NOBr( g) ⇌ 2NO( g) + Br 2( g) The change in concentration of [NO] will depend on the size of the vessel. The concentration of NO will increase. The concentration of NO will decrease. A catalyst will be needed to make a change in concentration. There will be no change in [NO].
Chemistry
1 answer:
jeyben [28]3 years ago
7 0

Answer:

The correct answer is : 'the concatenation of NO will increase'.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs.

2NOBr(g)\rightleftharpoons 2NO(g) + Br_2(g)

As, this is an endothermic reaction, increasing temperature will add more heat to the system which move equilibrium in the forward reaction with decrease in temperature. Hence, the equilibrium will shift in the right direction.

So, the concatenation of NO will increase.

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Answer : The image is attached below.

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Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n = \frac{m}{M}=\frac{24g}{48g/mol}=0.5mol

Number of particles, N = n\times 6.022\times 10^{23}=0.5\times 6.022\times 10^{23}=3.0\times 10^{23}

For NH_3:

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n = \frac{m}{M}=\frac{170g}{17g/mol}=10mol

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Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n = \frac{m}{M}=\frac{38g}{38g/mol}=1mol

Number of particles, N = n\times 6.022\times 10^{23}=1\times 6.022\times 10^{23}=6.0\times 10^{23}

For CO_2:

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m = n\times M=0.10mol\times 44g/mol=4.4g

Number of particles, N = n\times 6.022\times 10^{23}=0.10\times 6.022\times 10^{23}=6.0\times 10^{22}

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Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m = n\times M=0.20mol\times 46g/mol=9.2g

Number of particles, N = n\times 6.022\times 10^{23}=0.20\times 6.022\times 10^{23}=1.2\times 10^{23}

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Molar mass, M = 20 g/mol

Number of particles = 1.5\times 10^{23}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.5\times 10^{23}}{6.022\times 10^{23}}=0.25mol

Mass, m = n\times M=0.25mol\times 20g/mol=5g

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Molar mass, M = 44 g/mol

Number of particles = 1.2\times 10^{24}

Moles, n = \frac{N}{6.022\times 10^{23}}=\frac{1.2\times 10^{24}}{6.022\times 10^{23}}=1.9mol

Mass, m = n\times M=1.9mol\times 44g/mol=83.6g

For unknown substance:

Number of particles = 3.0\times 10^{23}

Mass, m = 8.5 g

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