At 2000 K, K_c = 0.154 for the reaction below. 2 CH_4(g) - C_2H_2(g) + 3 H_2(g)

Q: At 2000 K, K_c = 0.154 for the reaction below. 2 CH_4(g) - C_2H_2(g) + 3 H_2(g)

a) molefraction of H2 = nH2/(nCH4+nH2+nC2H2) = 0.6/(0.6+0.6) = 0.5 molefraction of H2 = 5*10^-1 B) If volume decreases,the pressure in the container increases.in order to decrease pressure equilibrium shifts towards less No of mole.so that No of moles of H2 decreaes.

ID: 1008067 • Letter: A

Question

At 2000 K, K_c = 0.154 for the reaction below. 2 CH_4(g) - C_2H_2(g) + 3 H_2(g) If a 1,25-L equilibrium mixture at 2000 K contains 0.60 mol each of CH_4(g) and H_2(g), answer the following questions. What is the mole fraction of H_2 present? Enter your answer with 2 significant digits. Enter scientific notation as 1.23E4. Enter only numbers in your answer. (Do not include units.) The mole fraction of H_2 is Number If the equilibrium mixture at 2000 K is transferred from a 1,25-L flask to 0.250-L flask, will the number of moles of H_2(g) increase, decrease, or remain unchanged?

Explanation / Answer

a) molefraction of H2 = nH2/(nCH4+nH2+nC2H2)

= 0.6/(0.6+0.6) = 0.5

molefraction of H2 = 5*10^-1

B) If volume decreases,the pressure in the container increases.in order to

decrease pressure equilibrium shifts towards less No of mole.so that No of

moles of H2 decreaes.

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At 2000 K, Kc = 0.154 for the reaction below. 2 CH4(g) C2H2(g) + 3 H2(g) If a 0.

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At 2000 K, K_c = 0.154 for the reaction below. 2 CH_4(g) - C_2H_2(g) + 3 H_2(g)

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