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Navami Ramson
Asked: 2022-11-08T01:24:11+05:30 In:

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :
A. `-245.7kJ//mol`
B. `-244.452kJ//mol`
C. `-241.5kJ//mol`
D. None of these

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :
A. `-245.7kJ//mol`
B. `-244.452kJ//mol`
C. `-241.5kJ//mol`
D. None of these
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  1. Correct Answer – D
    `H_(2)C_(2)O_(4)(l)+(1)/(2)O_(2)(g)rarrH_(2)O(l)+2CO_(2)(g)`,
    `Deltan_(g)=3//2`
    `DeltaU_(c )=-(0.312xx8.75)/(1)xx90`
    =-245.7 kJ/mol
    `DeltaH=DeltaU+Deltan_(g)RT`
    `=-245.7 +(3)/(2)xx(8.314xx300)/(1000)`
    =-241.95 kJ/mol

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Gajendra Chaudry
Asked: 2022-11-05T04:33:14+05:30 In:

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :
A. `-245.7KJ//mol`
B. `-244.452 KJ//mol`
C. `-246.947 KJ//mol`
D. none of these

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :
A. `-245.7KJ//mol`
B. `-244.452 KJ//mol`
C. `-246.947 KJ//mol`
D. none of these
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  1. Correct Answer – d

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Swati Chandran
Asked: 2022-10-29T17:09:04+05:30 In:

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :
A. `-245.7 KJ//mol`
B. `-244.452 KJ//mol`
C. `-246.947 KJ//mol`
D. None of these

When 1.0 g of oxalic acid `(H_(2)C_(2)O_(4))` is burnt in a bomb calorimeter whose capacity is 8.75 KJ/K, the enthalpy of combustion of oxalic acid at`27^(@)C` is :
A. `-245.7 KJ//mol`
B. `-244.452 KJ//mol`
C. `-246.947 KJ//mol`
D. None of these
Parabola
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1 Answer

  1. Correct Answer – C
    `H_(2)C_(2)O(l) + (1)/(2)O_(2)(g) rarr H_(2)O(l) + 2O_(2)(g),`
    `Deltan_(g)=3//2`
    `DeltaU_(0) =-(0.312 xx8.75)/(1) xx 90 =-245.7 KJ//”mol”`
    `DeltaH=DeltaU+ Deltan_(g)RT =- 245.7 KJ//”mol”`
    `DeltaH=DeltaU + Deltan_(g)RT =-245.7 +(3)/(2) xx (8.314xx300)/(1000)=-241.947 KJ//”mol”`

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