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Airida [17]
3 years ago
8

For the following reaction, Kc = 15 at 700 K. 2 NO(g) + Cl2(g) ⇄ 2 NOCl(g) If we have [NO] = 0.15 M, [Cl2] = 0.15 M, [NOCl] = 0.

40 M at 700 K, what will happen? Group of answer choices The equilibrium will not shift. The equilibrium will shift to the left, but will use up only part of the NOCl. The equilibrium will shift to the right, but will use up only part of the NO and Cl2. The equilibrium will shift to the right until all the reactants are used up. The equilibrium will shift to the left until all the NOCl is used up.
Chemistry
1 answer:
vovangra [49]3 years ago
6 0

Answer : The correct option is, The equilibrium will shift to the left until all the NOCl is used up.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

2NO(g)+Cl_2(g)\rightarrow 2NOCl(g)

The expression for reaction quotient will be :

Q=\frac{[NOCl]^2}{[NO]^2[Cl_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(0.40)^2}{(0.15)^2(0.15)}=47.4

The given equilibrium constant value is, K=15

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q>K that means product > reactant. So, the reaction is reactant favored that means reaction must shift to the reactants to be in equilibrium or shift to the left side.

Hence, the correct option is, The equilibrium will shift to the left until all the NOCl is used up.

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the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal
Sunny_sXe [5.5K]

Rydberg formula is given by:

\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )

where, R_{H} = Rydberg  constant = 1.0973731568508 \times 10^{7} per metre

\lambda = wavelength

n_{1} and n_{2} are the level of transitions.

Now, for n_{1}= 2 and n_{2}= 6

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )

= 1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.0278 )

= 1.0973731568508 \times 10^{7} \times 0.23

= 0.2523958\times 10^{7}

\lambda = \frac{1}{0.2523958\times 10^{7}}

= 3.9620\times 10^{-7} m

= 396.20\times 10^{-9} m

= 396.20 nm

Now, for n_{1}= 2 and n_{2}= 5

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )

= 1.0973731568508 \times 10^{7} \times (0.25-0.04 )

= 1.0973731568508 \times 10^{7} \times (0.21 )

= 0.230 \times  10^{7}

\lambda= \frac{1}{0.230 \times 10^{7}}

= 4.3478 \times 10^{-7} m

= 434.78\times 10^{-9} m

= 434.78 nm

Now, for n_{1}= 2 and n_{2}= 4

\frac{1}{\lambda} = 1.0973731568508 \times  10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.0625 )

= 1.0973731568508 \times 10^{7} \times (0.1875 )

= 0.20575 \times 10^{7}

\lambda= \frac{1}{0.20575 \times 10^{7}}

= 4.8602 \times 10^{-7} m

= 486.02 \times 10^{-9} m

= 486.02 nm

Now, for n_{1}= 2 and n_{2}= 3

\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )

=  1.0973731568508 \times 10^{7} \times (0.25-0.12 )

=  1.0973731568508 \times 10^{7} \times (0.13 )

= 0.1426585\times 10^{7}

\lambda= \frac{1}{0.1426585\times 10^{7}}

= 7.0097 \times 10^{-7} m

= 700.97 \times 10^{-9} m

= 700.97 nm



5 0
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Answer:

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ivann1987 [24]

Answer:

C.

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Looking across period 4 of the periodic table, potassium (atomic number 19) is followed by calcium (atomic number 20), which is
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So the element on the left most will have the largest atomic radius. So the correct ans is Potassium. Potassium will have the largest atomic size among Potassium, Calcium and Scandium. 
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