Question

12). Given the two equations: 2 C(s) + O2(8) 2 co(e) Ke=0.085 at 450 R CHOH(E)=CO+ 2 H (g). Kc = 62 at 450 R a) (10 pts) Find
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Answer #1

Part a)

cis) - ola) Ке, - [о] [0,1 2 со (9) ka-oogs (0) Ke, Co,) : [о] - - 2ң, (9) ke, - 6 он (9) — со ke, Гео]4,7- (оң он] Reauіrеd

(There is a change in required equation as the given equation is not balanced in hydrogen. Balanced equation has been used in answer).

Part b)

Moles of CH3OH = 1 g/32 g/mol = 0.03125 mol

[CH3OH ] = 0.03125 mol/5 L = 0.00625 M

[O2] = (1 g/32 g/mol) / 5 L = 0.00625 M

[H2] = (1 g/2 g/mol) / 5 L = 0.100 M

Qc = [CH3OH]2/[O2][H2]4

= (0.00625)2/(0.00625)(0.100)4

= 62.5

As Qc is more than Kc, reaction will proceed towards reactants.

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12). Given the two equations: 2 C(s) + O2(8) 2 co(e) Ke=0.085 at 450 R CHOH(E)=CO+ 2 H (g). Kc = 62 at 450 R a) (10...
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