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Diketahui reaksi : 2C 2 H 6 (g) + 7O 2 (g) → 4CO 2 (g) + 6 H 2 O(l) ∆H = -3130 kJ H 2 (g) + ½O 2 (g) H 2 O(l) ∆H = -286 kJ C 2 H 2 (g) + 2H 2 (g) C 2 H 6 (g) ∆H = -312 kJ Berapa ∆H yang dibebaskan jika 11,2 liter gas C 2 H 2 dibakar sempurna pada keadaan standar?

Diketahui reaksi :

2C2H6 (g) + 7O2(g)  4CO2(g) + 6 H2O(l)       ∆H = -3130 kJ

H2(g) + ½O2(g) rightwards arrow H2O(l)                                 ∆H = -286 kJ

C2H2(g) + 2H2(g) rightwards arrow C2H6(g)                            ∆H = -312 kJ

Berapa ∆H yang dibebaskan jika 11,2 liter gas C2H2 dibakar sempurna pada keadaan standar?

  1. +652,5 kJ

  2. -652,5 kJ

  3. +1864 kJ

  4. -1864 kJ

  5. +1305 kJ

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Pembahasan

2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6 H 2 O(l) ∆H = -3130 kJ (tetap) H 2 (g) + ½O 2 (g) H 2 O(l) ∆H = -286 kJ (dibalik, kali 4) C 2 H 2 (g) + 2H 2 (g) C 2 H 6 (g) ∆H = -312 kJ (tetap, kali 2) Maka reaksinya menjadi : 2C 2 H 6 (g) + 7O 2 (g) 4CO 2 (g) + 6 H 2 O(l) ∆H = -3130 kJ 4H 2 O(l) 4H 2 (g) + 2O 2 (g) ∆H = -286 kJ x 4 = +1144 kJ 2C 2 H 2 (g) + 4H 2 (g) 2C 2 H 6 (g) ∆H = -312 kJ x2 = -624 kJ + 5O 2 2H 2 O Jadi, reaksi pembakaran 2 mol asitilena melepaskan 2610 kJ kalor *mol 11,2 Liter asitilena (C 2 H 2 ) dalam keadan standar Mol = *entalpi untuk 0,5 mol asitilena 2x = -1305 kJ X = -652, 5 kJ

2C2H6 (g) + 7O2(g) rightwards arrow 4CO2(g) + 6 H2O(l)       ∆H = -3130 kJ (tetap)

H2(g) + ½O2(g) rightwards arrow H2O(l)                                 ∆H = -286 kJ (dibalik,  kali 4)

C2H2(g) + 2H2(g) rightwards arrow C2H6(g)                            ∆H = -312 kJ (tetap, kali 2)

Maka reaksinya menjadi :

 

2C2H6 (g) + 7O2(g) rightwards arrow 4CO2(g) + 6 H2O(l)       ∆H = -3130 kJ

4H2O(l) rightwards arrow 4H2(g) + 2O2(g)                              ∆H = -286 kJ x 4 = +1144 kJ

2C2H2(g) + 4H2(g) rightwards arrow 2C2H6(g)                        ∆H = -312 kJ x2 = -624 kJ       +

                                  5O2                                                 2H2O

up diagonal strike 2 C 2 H 6 space left parenthesis g right parenthesis end strike space plus space up diagonal strike 7 O 2 left parenthesis g right parenthesis end strike space space rightwards arrow 4 C O 2 left parenthesis g right parenthesis space plus space up diagonal strike 6 space H 2 O left parenthesis l right parenthesis end strike space space space increment H space equals space minus 3130 space k J up diagonal strike 4 H 2 O left parenthesis l right parenthesis end strike space space rightwards arrow up diagonal strike 4 H 2 left parenthesis g right parenthesis end strike space plus space up diagonal strike 2 O 2 left parenthesis g right parenthesis end strike space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space increment H space equals space minus 286 space k J space x space 4 space equals space plus 1144 space k J fraction numerator 2 C 2 H 2 left parenthesis g right parenthesis space plus space up diagonal strike 4 H 2 left parenthesis g right parenthesis end strike space space rightwards arrow up diagonal strike 2 C 2 H 6 left parenthesis g end strike right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space increment H space equals space minus 312 space k J space x 2 space equals space minus 624 space k J space space space space space space space plus over denominator 2 C 2 H 2 left parenthesis g right parenthesis space plus space 5 O 2 left parenthesis g right parenthesis rightwards arrow ​ space 4 C O 2 left parenthesis g right parenthesis space plus space 2 H 2 O left parenthesis l right parenthesis space space space space space space space space space increment H space equals space minus 2610 space k J end fraction

Jadi, reaksi pembakaran 2 mol asitilena melepaskan 2610 kJ kalor

 

*mol 11,2 Liter asitilena (C2H2) dalam keadan standar

Mol =  fraction numerator 11 comma 2 space l i t e r over denominator 22 comma 4 space l i t e r end fraction equals 0 comma 5 space m o l

 

*entalpi untuk 0,5 mol asitilena

fraction numerator 2 space space m o l over denominator 0 comma 5 space m o l end fraction equals fraction numerator negative 2610 space k J over denominator x end fraction 2 x equals space minus 1305 space k J x equals space minus 652 comma 5 space k J

 

2x = -1305 kJ

X = -652, 5 kJ

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