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Lostsunrise [7]
3 years ago
14

Coal can be used to generate hydrogen gas (a potential fuel) by thefollowing endothermic reaction.C(s) + H2O (g) <==> CO(g

) + H2(g)if this reaction mixture is at equilibrium, predict whether each ofthe following will result in the formation of additional hydrogengas, the formation of less hydrogen gas, or have no effect on thequantity og hydrogen gas.a. adding more C to the reaction mixtureb. adding more H2O to the reaction mixturec. raising the temperature of the reaction mixtured. increasing the volume of the reaction mixturee. adding a catalyst to the reaction mixturef. adding an inert gas to the reaction mixture
Chemistry
1 answer:
SpyIntel [72]3 years ago
6 0

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.

For the given equation:

C(s) + H_2O (g)\leftrightharpoons CO(g) + H_2(g)

Given that reaction is an endothermic reaction.

For the given options:

a)Adding more C

If the concentration of C that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of C takes place. Therefore, the equilibrium will shift in the right direction to wards the formation of hydrogen gas.

b) Adding more H_2O

If the concentration of water that is the reactant is increased, so according to the Le-Chatlier's principle, the equilibrium will shift in the direction where decrease of concentration of water  takes place. Therefore, the equilibrium will shift in the right direction towards the formation of hydrogen gas.

c) Raising the temperature  of the reaction mixture

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

As, this is an endothermic reaction, forward reaction will decrease the temperature. Hence, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.  

d) Increasing the volume  of reaction mixture

If the volume of the container is increased, the pressure will decrease according to Boyle's Law. Now, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where increase in pressure is taking place. As the number of moles of gas molecules is greater at the product side. So, the equilibrium will shift in the right direction that is towards the formation of hydrogen gas.  

e) Adding a catalyst  to reaction mixture

Role of catalyst is to attain the equilibrium quickly without disturbing the state of equilibrium. Hence, addition of catalyst will not change the equilibrium of the reaction.  

f) Adding an inert gas to reaction mixture

Adding inert gas to the mixture at constant volume will not effect the equilibrium. Hence, addition of an inert gas will not change the equilibrium of the reaction.  

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Answer:

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

Explanation:

Given :

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula:

E=\frac{R_y}{n^2}

R_y=2.18\times 10^{-18} J =  Rydberg energy

n =  principal quantum number of the orbital

Energy of 11th orbit = E_{11}

E_{11}=\frac{2.18\times 10^{-18} J}{11^2}=1.80\times 10^{-20} J

Energy of 10th orbit = E_{10}

E_{10}=\frac{2.18\times 10^{-18} J}{10^2}=2.18\times 10^{-20} J

Energy difference between both the levels will corresponds to the energy of the wavelength of the line which can be calculated by using Planck's equation.

E'=E_{10}-E_{11}=2.18\times 10^{-20} J-1.80\times 10^{-20} J

=E'=0.38\times 10^{-20} J

\lambda =\frac{hc}{E'} (Planck's' equation)

\lambda = \frac{6.626\times 10^{-34} Js\times 3\times 10^8 m/s}{0.38\times 10^{-20} J}

\lambda = 5.2310\times 10^{-5} m\approx 5.23\times 10^{-5} m

The wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron for the given energy levels is 5.23\times 10^{-5} m

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