Question

# 0.16 g of methane was subjected to combustion at $$\displaystyle 27^{\circ}$$ C in a bomb Calorimeter. The temperature of calorimeter system (including water) was found to rise by $$\displaystyle 0.5^{\circ}$$ C. The heat of combustion of methane at :( i ) constant volume and ( ii ) constant pressure.[Given: The thermal capacity of calorimeter system is 17.7 kJ $$\displaystyle K^{-1}.$$ $$\displaystyle \left ( R=8.314\: J\: mol^{-1}K^{-1} \right )]$$

A
( i ) 885 kJ / mol ( ii ) 889.95 kJ / mol
B
( i ) 785 kJ / mol ( ii ) 859.95 kJ / mol
C
( i ) 587kJ / mol ( ii ) 789.95 kJ / mol
D
( i ) 985 kJ / mol ( ii ) 999.95 kJ / mol

Solution

## The correct option is B ( i ) $$\displaystyle -$$885 kJ / mol ( ii ) $$\displaystyle -$$889.95 kJ / mol$${ CH }_{ 4 }\left( g \right) +2{ O }_{ 2 }\left( g \right) \rightarrow C{ O }_{ 2 }(g)+2{ H }_{ 2 }O(l)$$Heat of combustion at constant volume i.e., $${ q }_{ v }=\Delta E=$$ internal energy.$$\Delta E=$$ Heat capacity of calorimeter$$\times$$(rise in temperature)$$\times$$((molar mass of compound)/(mass of compound))        $$\displaystyle=17.7\times0.5\times\frac { 16 }{ 0.16 } =885$$$${ q }_{ { v } }=\Delta E=-885\ kJ/mol$$Heat of combustion at constant pressure $$={ q }_{ p }=\Delta H=$$ethalpy$$\Delta H=\Delta E+\Delta nRT$$$$\Delta n=1-3=-2$$$$T=300K;R=8.314\times{ 10 }^{ -3 }kJ.{ K }^{ -1 }/mol$$So, $$\Delta H=-885+(-2)\times (8.314\times{ 10 }^{ -3 })\times 300=-889.99\ kJ/mol$$Chemistry

Suggest Corrections

0

Similar questions
View More

People also searched for
View More