In a particular reaction, 4 text{ kJ} of heat | vedclass

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

Question

(B) Step $1$: Identify the sign conventions for heat $(q)$ and work $(w)$.Heat released by the system: $q = -4 	ext{ kJ}$.Work done on the system: $w

Vedclass · Accepted Answer

$\Delta H = -4 	ext{ kJ}$ and $\Delta U = 8 	ext{ kJ}$

Vedclass · Answer

(B) Step $1$: Identify the sign conventions for heat $(q)$ and work $(w)$.Heat released by the system: $q = -4 	ext{ kJ}$.Work done on the system: $w = +12 	ext{ kJ}$.Step $2$: Calculate the change in internal energy $(\Delta U)$ using the first law of thermodynamics: $\Delta U = q + w$.$\Delta U = -4 	ext{ kJ} + 12 	ext{ kJ} = 8 	ext{ kJ}$.Step $3$: For a reaction at constant pressure, the heat released is equal to the change in enthalpy $(\Delta H)$.$\Delta H = q_p = -4 	ext{ kJ}$.Thus, $\Delta H = -4 	ext{ kJ}$ and $\Delta U = 8 	ext{ kJ}$.

In a particular reaction, $4 \text{ kJ}$ of heat is released by the system and $12 \text{ kJ}$ of work is done on the system. Calculate the $\Delta H$ and $\Delta U$.

Similar Questions

$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$. The enthalpy change $(\Delta H)$ for the above reaction at $27^{\circ}C$ is $-1366.5 \text{ kJ mol}^{-1}$. Calculate the internal energy change $(\Delta U)$ for the same reaction at this temperature.

For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ at constant temperature and pressure,if $\Delta H$ and $\Delta U$ are the enthalpy and internal energy changes for the reaction,which of the following equations is correct?

Work done in vaporisation of one mole of water at $373 \ K$ and the pressure of $1 \ atm$ is approximately (for ideal gas)......$J$

Two moles of an ideal monoatomic gas are allowed to expand adiabatically and reversibly from $300 \ K$ to $200 \ K$. The work done in the system is $... \ kJ$ $(C_V = 12.5 \ J/K/mol)$.

Assertion : For a reaction $2NH_{3(g)} \to N_{2(g)} + 3H_{2(g)}$ ; $\Delta H > \Delta E$.
Reason : Enthalpy change is always greater than internal energy change.

Vedclass Test Series

Exam Paper Generator

Online Exam Module