CH3COOH(g) can be prepared using equation CO(g) + 2H2(g)⇌CH3COOH(g), KC=4×102.At equilibrium, a 5L flask contains equal moles of CO(g) and CH3COOH(g). Hence, number of moles of hydrogen at equilibrium:
0.25mole
0.10mole
0.50mole
0.125mole
A
0.10mole
B
0.125mole
C
0.50mole
D
0.25mole
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Q1
CH3COOH(g) can be prepared using equation CO(g) + 2H2(g)⇌CH3COOH(g), KC=4×102.At equilibrium, a 5L flask contains equal moles of CO(g) and CH3COOH(g). Hence, number of moles of hydrogen at equilibrium:
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Q2
CH3OH(g) can be prepared using equation, CO(g)+2H2(g)⇌CH2OH(g),Kc=4×102 At equilibrium, a 5 L flask contains equal moles of CO(g) and CH3OH(g). Hence, number of moles of H2 at equilibrium is:
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Q3
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At eqbm. 5L flask contains equal moles of CO and CH3OH . Hence the no. Of moles of H2 at eqbm is
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