In a 1 , L container,2 , mol of CO and 1 , mo | vedclass

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(C) The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta(PV)$.Given $\Delta H = -560 \, kJ$.We need to find $

Vedclass · Accepted Answer

$-557$

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(C) The relationship between $\Delta H$ and $\Delta U$ is given by: $\Delta H = \Delta U + \Delta(PV)$.Given $\Delta H = -560 \, kJ$.We need to find $\Delta(PV) = P_2V_2 - P_1V_1$.Initial state: $P_1 = 70 \, atm$,$V_1 = 1 \, L$,so $P_1V_1 = 70 \, atm \cdot L$.Final state: $P_2 = 40 \, atm$,$V_2 = 1 \, L$,so $P_2V_2 = 40 \, atm \cdot L$.Thus,$\Delta(PV) = (40 - 70) \, atm \cdot L = -30 \, atm \cdot L$.Using the conversion factor $1 \, atm \cdot L = 0.1 \, kJ$,we get $\Delta(PV) = -30 	imes 0.1 \, kJ = -3 \, kJ$.Substituting into the equation: $-560 \, kJ = \Delta U + (-3 \, kJ)$.Therefore,$\Delta U = -560 + 3 = -557 \, kJ$.

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