The relationship between Δ H and Δ U for the reaction, PCl_3(g) + Cl_2(g) arrow PCl_5(g) is given as

Correct answer: D. Δ H = Δ U - RT The given reaction is PCl_3(g) + Cl_2(g) arrow PCl_5(g) The relationship between enthalpy change (Δ H ) and internal energy change (Δ U ) is given by the equation: Δ H = Δ U + Δ n_g RT where Δ n_g is the change in the number of moles of gaseous species. Δ n_g = moles of gaseous products - moles of gaseous reactants Δ n_g = 1 - (1 + 1) = -1 Substituting Δ n_g = -1 in the equation, we get: Δ H = Δ U - RT Answer: Δ H = Δ U - RT

MHT CET 202626 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 26 April
Shift: Morning Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

MHT CET 2026 · 26 April · Morning Shift · Chemistry · Thermodynamics (C)

The relationship between Δ H and Δ U for the reaction, PCl_3(g) +...

The relationship between

\Delta H

and

\Delta U

for the reaction,

\text{PCl}_3(g) + \text{Cl}_2(g) \rightarrow \text{PCl}_5(g)

is given as

Options

A $\Delta H = \Delta U + 2RT$
B $\Delta H = \Delta U - 2RT$
C $\Delta H = \Delta U + RT$
D $\Delta H = \Delta U - RT$

Correct answer

D. $\Delta H = \Delta U - RT$

Step-by-step solution

The given reaction is

\text{PCl}_3(g) + \text{Cl}_2(g) \rightarrow \text{PCl}_5(g)

The relationship between enthalpy change (\Delta H

) and internal energy change (\Delta U

) is given by the equation:

\Delta H = \Delta U + \Delta n_g RT

where

\Delta n_g

is the change in the number of moles of gaseous species.

\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}

\Delta n_g = 1 - (1 + 1) = -1

Substituting

\Delta n_g = -1

in the equation, we get:

\Delta H = \Delta U - RT

Answer:

\Delta H = \Delta U - RT

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