The standard enthaply change H degrees for the thermal decomposition of silver n

Question

The standard enthaply change H degrees for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s)--> AgNO2(s) + 1/2O2(g). The standard enthaply of formation of AgNO3(s) is -123.02 kJ/mol. calculate the standard enthaply of formation of AgNO2(s). (kJ/mol) The standard enthaply change H degrees for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s)--> AgNO2(s) + 1/2O2(g). The standard enthaply of formation of AgNO3(s) is -123.02 kJ/mol. calculate the standard enthaply of formation of AgNO2(s). (kJ/mol) The standard enthaply change H degrees for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s)--> AgNO2(s) + 1/2O2(g). The standard enthaply of formation of AgNO3(s) is -123.02 kJ/mol. calculate the standard enthaply of formation of AgNO2(s). (kJ/mol) The standard enthaply change H degrees for the thermal decomposition of silver nitrate according to the following equation is 78.67 kJ: AgNO3(s)--> AgNO2(s) + 1/2O2(g). The standard enthaply of formation of AgNO3(s) is -123.02 kJ/mol. calculate the standard enthaply of formation of AgNO2(s). (kJ/mol)

Explanation / Answer

AgNO3(s)--> AgNO2(s) + 1/2O2(g) H = 78.67 kJ

HRxn = Hrpod - Hreact

if Hreact = -123.02 kJ/mo

and Hf O2 = 0 then

HRxn = Hrpod - Hreact

78.67 kJ = (AgNO2(s)) -  -123.02

123.02+78.67 = (AgNO2(s))

H (AgNO2(s)) = 201.69 kJ/mol

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