At constant T and P, which one of the following statements is correct for the reaction, $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$

Δ H

is independent of the physical state of the reactants of that compound

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ H > Δ E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Δ H < Δ E

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Δ H = Δ E

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $Δ H < Δ E$
$Δ n_{g} = 1 - \frac{3}{2} = \frac{- 1}{2}, A s Δ n_{g}$ is negative, thus $Δ H < Δ E$

Suggest Corrections

Similar questions

At constant temperature and pressure, a correct statement for the reaction : $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ is:

Which of the following statements is correct for the reaction $C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)$ at constant temperature and pressure?

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g)

C O (g) + \frac{1}{2} O_{2} (g) \to {C O}_{2} (g)

C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H_{1} = a

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H_{2} = b

C (s) + C O_{2} (g) \to 2 C O (g); Δ H_{3} = c

Join BYJU'S Learning Program