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The heat envoled in the combustion of gl...

The heat envoled in the combustion of glucose is given by the equation
`C_(6)H_(12)O_(6(s))+6O_(2(g))rarr6CO_(2(g))+6H_(2)O_((g))`
`DeltaH=-680K.cal,` The weight of `CO_(2(g))` produced when 170 Kcal of heat is envolved in the combusation of glucose is

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To find the weight of \( CO_2 \) produced when 170 Kcal of heat is evolved in the combustion of glucose, we can follow these steps: ### Step 1: Understand the Reaction The combustion of glucose can be represented by the equation: \[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(g) \] The heat evolved (\( \Delta H \)) for this reaction is given as -680 Kcal. ...
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The value of x in balanced equation is C_(6)H_(12)O_(6(aq))+6O_(2(g))to"x"CO_(2(g))+6H_(2)O_((l))+"energy"

The heat evolved in the combustion of glucose is shown in the equation : C_(6)H_(12)O_(6)(s) + 6O_(2)(g) rarr 6CO_(2)(g) + 6H_(2)O (g) , Delta_(c ) H = -2840 kJ mol^(-1) What is the energy requirement for production of 0.36g of glucose by the reverse reaction ?

Knowledge Check

  • If C_(6)H_(12)O_(6)(s)+9O_(2)(g)rarr6CO_(2)(g)+6H_(2)O(g) , Delta H=-680 Kcal The weight of CO_(2)(g) produced when 170 Kcal of heat is evolved in the combustion of glucose is :-

    A
    265 gm
    B
    66 gm
    C
    11 gm
    D
    64 gm

  • According to the equation C_(6)H_(6(l))+7/2 O_(2(g)) rarr 6CO_(2(g))+3H_(2)O_((l)), DeltaH=-xkJ The energy evolved when 3.9 gm of benzene is burnt in air is - 163.2 kJ heat of combustion of benzene is

    A
    32.46 kJ
    B
    16.32 kJ
    C
    326.4 kJ
    D
    `-3264 kJ`

  • The heat evolved in the combustion of benzene is given by the equation: C_(6)H_(6)(g) + (15)/(2) O_(2)(g) to 6CO_(2)(g) + 3H_(2)O(l), DeltaH = - 3264.6 kJ "mol"^(-1) The heat energy changes when 39 g of C_(6)H_(6) are burnt im an open ontainer will be:

    A
    `+ 816.15 kJ "mol"^(-1)`
    B
    `+ 1632.3 kJ "mol"^(-1)`
    C
    `-1632.3 kJ "mol"^(-1)`
    D
    `-2448.45 kJ "mol"^(-1)`

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