Question

How much heat (in kJ) is released when 15.0 L of CO at 85°C and 112 kPa reacts with 14.4 L of H2 at 75°C and 744 torr?

Answer 1

Heat energy released is - 31.4 kJ.

Explanation:

CO₍g₎ + 2H₂₍g₎ → CH₃OH₍l₎

Volume of CO, V ₍CO₎ = 15 L = 0.015 m³

Pressure = 112 kPa = 112,000 Pa

T= 85 ⁰C = 85 + 273 = 358 K

As per ideal gas law,

PV = nRT

n = $\frac{PV}{RT}$

   = 112000 × 0.015 / 8.314 × 358

 n(CO) = 0.56 moles

Volume of H₂ = 14.4 L = 0.0144 m³

Pressure = 744 torr = 99191.84 Pa

T= 75⁰C + 273 K = 348 K

n(H₂) = 99191.84 ×0.0144 m³ / 8.314 ×348 K

     = 0.49 moles of H₂

The above calculation shows that hydrogen is the limiting reagent.

n(CH₃OH) = n(H₂) /2

        = 0.49/2 = 0.245 mol

ΔH₍rxn₎ = ΔH₍f₎ (CH₃OH) - ΔH₍f₎(CO)

           = -238.6 -(-110.5)

           = -128.1 kJ/mol

Now we have to multiply ΔH₍rxn₎ with the moles of methanol.

E = ΔH₍rxn₎ × n(CH₃OH) = -128.1 ×0.245

        = - 31.4 kJ.