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

Question

At 2000 K, Kc = 0.154 for the reaction below. 2 CH4(g) C2H2(g) + 3 H2(g) If a 0.750-L equilibrium mixture at 2000 K contains 0.45 mol each of CH4(g) and H2(g), answer the following questions. (a) What is the mole fraction of C2H2 present? (Note: The mole fraction of C2H2 is equal to the number of moles of C2H2 divided by the total number of moles.) 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 C2H2 is ____________ (b) If the equilibrium mixture at 2000 K is transferred from a 0.750-L flask to 10.0-L flask, will the number of moles of C2H2(g) increase, decrease, or remain unchanged? __________

Explanation / Answer

M = mol/V so:

[CH4] = 0.45/0.75 = 0.6

[H2] = 0.6

[C2H2] = x

solve for x

Kc = [C2H2][H2]^3/[CH4]^2

0.154 = (x)(0.6^3)/(0.6^2)

x =0.154 /0.6 = 0.2566

so

[C2H4] = 0.2566M

mol fraction = 0.2566 / (0.2566+ 0.6 +0.6) = 0.176163

mol frac. C2H4 = 0.18

b)

if eq. is V = 0.75 L --> 10L

this will favour, the most mol formation, i.e. products, since

2 mol of gas --> 4 mol of gas

so there is more space for products

so it will INCREASE

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