8) The enthalpy change for the following reaction is -483.6 k: 2H2 (g) + O2 (g)
ID: 875241 • Letter: 8
Question
8) The enthalpy change for the following reaction is -483.6 k: 2H2 (g) + O2 (g) 2H2O (g) Therefore, the enthalpy change for the following reaction isk kJ: 4H2 (g) + 202 (g) 4H2O (g) D)-483.6 A) 967.2 B) -967.2 483.6 D) -483.6 9) The value of AH for the reaction below is +128.1 k) CH3OH (l) CO (g) + 2 H2 (g) How many kJ of heat are consumed when 15.5 g of CH3OH (1) decomposes as shown in the equation? A) 32 B) 0.48 9833D) 620 10) The value of AH for the reaction below is +128,1 kj CH3OH () CO() + 2H2(8) How many KJ of heat are consumed when 5.75 g of CO (8) is formed as shown in the equation? A) 26.3 B) 23.3 C) 83 D) 620
Explanation / Answer
8) option B) -967.2
Enthalpy change for the following reaction is -483.6kJ
2 H2 + O2 ------------------> 2 H2O
Enthalpy change for the following reaction is
4 H2 + 2 O2 ------------------> 4 H2O
Enthalpy change = 2 x -483.6 = -967.2 kJ
9) option D) 62.0
delta H value for the following reaction is 128.1 kJ
CH3OH -----------------------> CO + 2 H2
32.04g 28.01g 2 x 2g
32.04 g CH3OH decomposes ----------------------- 128.1 kJ heat consumed
15.5 g of CH3OH decomposes ------------------------- ? kJ heat consumed
heat consumed = 15.5 x 128.1 / 32.04
= 62.0 kJ
10) option A) 26.3
128.1 kJ is consumed ---------------to form 28.01 g of CO
how many kJ is consumed----------------- to form 5.75 g of CO
heat consumed = 5.75 x 128.1 / 28.01
= 26.3 kJ
11) answer : D ) -20.8 kJ
2Ba + O2 -------------->2BaO
2x153.2 =306.4
306.4g BaO ---------------> -1107 kJ
5.75 g BaO ----------------> ?
heat = 5.75 x -1107/306.4
= - 20.8 kJ
12 ) answer : A) 1.13 J/g-K
just divide the 88 with molar mass of C2H6SO
13 ) answer : D ) 0.13 J/g 0C
Q= m x Cp x (T2-T1)
29 =15 x Cp x (37-22)
Cp = 0.128
= 0.13
14) ansewr : A ) 16.2
mass = d x v
= 3.12 x 10
= 31.2
Q = m x Cp x (T2-T1)
= 31.2 x 0.226 x (27.3-25)
= 16.2 J
15) answer : A ) +355 kJ
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