Bài tập Tính biến thiên enthalpy của phản ứng theo năng lượng liên kết lớp 10 (có lời giải)

A. +158 kJ.                   

B. -158 kJ.

C. +185 kJ.                   

D. -185 kJ.

A. -560 kJ.                    

B. -506 kJ.

C. +560 kJ.                   

D. +506 kJ.

A. -3734 kJ.                     

B. +3734 kJ.

C. -46 kJ.                         

D. +46 kJ.

A. +103 kJ.                       

B. – 103 kJ.      

C. +80 kJ.                           

D. – 80 kJ.

A. +134.                                 

B. -134.

C. +478.                                 

D. +284.

A. -169 kJ.                    

B. +169 kJ.

C. -196 kJ.                    

D. +196 kJ.

A. 305.                                   

B. 360.

C. 464.                                   

D. 837.

A. \[{\Delta _r}H_{298K}^o = m \times {\Delta _f}H_{298}^o\left( M \right) + n \times {\Delta _f}H_{298}^o\left( N \right) + a \times {\Delta _f}H_{298}^o\left( A \right) + b \times {\Delta _f}H_{298}^o\left( B \right)\]

B. \[{\Delta _r}H_{298K}^o = a \times {\Delta _f}H_{298}^o\left( A \right) + b \times {\Delta _f}H_{298}^o\left( B \right) - m \times {\Delta _f}H_{298}^o\left( M \right) - n \times {\Delta _f}H_{298}^o\left( N \right)\]

C. \[{\Delta _r}H_{298K}^o = a \times {E_b}\left( A \right) + b \times {E_b}\left( B \right) - m \times {E_b}\left( M \right) - n \times {E_b}\left( N \right)\]

D. \[{\Delta _r}H_{298K}^o = m \times {E_b}\left( M \right) + n \times {E_b}\left( N \right) - a \times {E_b}\left( A \right) - b \times {E_b}\left( B \right)\]

A. +105 kJ.                   

B. -105 kJ.

C. +150 kJ.                   

D. -150 kJ.

A. \[{\Delta _r}H_{298}^0 = {\Delta _f}H_{298}^0\left( {{C_2}{H_5}OH} \right){\rm{ + }}{\Delta _f}H_{298}^0\left( {{C_2}{H_4}} \right){\rm{ + }}{\Delta _f}H_{298}^0\left( {{H_2}O} \right)\]

B. \[{\Delta _f}H_{298}^0 = {\Delta _r}H_{298}^0\left( {{C_2}{H_5}OH} \right){\rm{ }} - {\Delta _r}H_{298}^0\left( {{C_2}{H_4}} \right){\rm{ }} - {\Delta _r}H_{298}^0\left( {{H_2}O} \right)\]

C. \[{\Delta _r}H_{298}^0 = {\Delta _f}H_{298}^0\left( {{C_2}{H_5}OH} \right){\rm{ }} - {\Delta _f}H_{298}^0\left( {{C_2}{H_4}} \right){\rm{ }} - {\Delta _f}H_{298}^0\left( {{H_2}O} \right)\]

D. \[{\Delta _r}H_{298}^0 = {\rm{ }}{\Delta _f}H_{298}^0\left( {{C_2}{H_4}} \right){\rm{ + }}{\Delta _f}H_{298}^0\left( {{H_2}O} \right) - {\Delta _f}H_{298}^0\left( {{C_2}{H_5}OH} \right)\]