Question

please show work

An industrial chemist introduces 2.0 atm of H₂ and 2.0 atm of CO₂ into a 1.00−L container at 25.0°C and then raises the temperature to 700°C, at which K_c = 0.534:

H₂(g) + CO₂(g) ⇌ H₂O (g) + CO(g)

How many grams of H₂ are present at equilibrium?

Answer 1

We need to apply multiple concepts here.

Increasing the temperature from 25^oC (298 K) to 700^oC (973 K) will increase the pressure of each reactant gases :

Let x be the change in pressure of reactants when they reach equilibrium.

P(H₂) = 6.53 - x
P(CO₂) = 6.53 - x
P(H₂O) = x
P(CO) = x

Hence, equilibrium expression is given by:

Solving for X, we get X = 2.76 atm

Therefore, Pressure of H₂ at equilibrium =  6.53 - 2.76 = 3.77 atm

From ideal gas equation, no.of moles of H₂ is given by:

Therefore, Mass of H₂ at equilibrium = (0.0472 mol) x( 2 g/mol) = 0.0944 g of H₂

• Mass of H₂ at equilibrium = 0.0944 g

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