For the following reaction,the relation betwe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \De

Vedclass · Accepted Answer

$\Delta H = \Delta U - RT$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + \Delta n_g RT$.For the reaction: $2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$.The change in the number of moles of gaseous species is calculated as: $\Delta n_g = \sum n_{p(g)} - \sum n_{r(g)}$.$\Delta n_g = 2 - (2 + 1) = 2 - 3 = -1$.Substituting this value into the equation,we get: $\Delta H = \Delta U + (-1) RT$.Therefore,$\Delta H = \Delta U - RT$.

For the following reaction,the relation between $\Delta H$ and $\Delta U$ is:
$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$

Similar Questions

One mole of an ideal gas at $300 \ K$ and $20 \ atm$ expands to $2 \ atm$ under isothermal and reversible conditions. The work done by the gas is $-x \ kJ \ mol^{-1}$. The value of $x$ is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

An ideal gas expands isothermally from $10^{-3} \; m^{3}$ to $10^{-2} \; m^{3}$ at $300 \; K$ against a constant external pressure of $10^{5} \; Nm^{-2}$. The work done on the gas is:

Five moles of an ideal gas at $1 \ bar$ and $298 \ K$ is expanded into vacuum to double the volume. The work done is :-

Choose the correct answer. $A$ thermodynamic state function is a quantity

When the temperature of $2$ moles of an ideal gas is increased by $20^{\circ} C$ at constant pressure,find the work involved in the process. (in $R$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For the following reaction,the relation between $\Delta H$ and $\Delta U$ is:$2 SO_{2(g)} + O_{2(g)} \rightarrow 2 SO_{3(g)}$