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

Vedclass · Accepted Answer

$-89.85$

Vedclass · Answer

(B) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta U)$ is given by the equation: $\Delta H = \Delta U + P\Delta V$. Given: $\Delta H = -92.2 \ kJ/mol$,$P = 20.0 \ atm$,and $\Delta V = -1.16 \ L$. First,calculate the work done $(P\Delta V)$ in $L \cdot atm$: $P\Delta V = 20.0 \ atm 	imes (-1.16 \ L) = -23.2 \ L \cdot atm$. Convert $L \cdot atm$ to $kJ$ using the conversion factor $1 \ L \cdot atm = 0.1013 \ kJ$: $P\Delta V = -23.2 	imes 0.1013 \ kJ = -2.35 \ kJ$. Now,substitute the values into the equation: $-92.2 \ kJ = \Delta U + (-2.35 \ kJ)$. $\Delta U = -92.2 \ kJ + 2.35 \ kJ = -89.85 \ kJ$.

For the reaction of nitrogen with hydrogen to form ammonia,$\Delta H^o = -92.2 \ kJ/mol$. If the reaction is carried out at a constant pressure of $20.0 \ atm$ and a volume change of $-1.16 \ L$,calculate the change in internal energy $(\Delta U)$ in $kJ$.

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