17. Calculate K, forHg)+0H,0) at 600 K, using the following data H2(g) +02(g) H2

Question

17. Calculate K, forHg)+0H,0) at 600 K, using the following data H2(g) +02(g) H2O2(g) K, 2H2(g) + 02(g) 2H20(g) 2.3 × 106 at 600 K K, .8 × 1037 at 600 K ICJ 5.4 × 10-13 [A] 9.8 × 1024 [B126x10-31 [DI 1.2 × 10- 4.4x 100 18. Given the equation: 2NOCIfg) at 115'C. Calculate Kp A] 141 x 104 B1 0.478 ICI 0.142 2N0g) + atg). The coluilibrium constant is 00150 [D) 0.0150 EJ none of these 19. For the reaction below, K, = 1.16 at 8000. If a 20.0-gram sample of CaCO, is put into a 10.0-liter container and heated to 800 C, what percent of the CaCO, will react to reach equilibrium? [A] 34.1% [B) 14.6% ICI 65.9% [DI 1000% IEI none of these 20. At-80°C. K for the reaction is 4.66 x 10-8 We introduce 0.050 mole of N 04 into a 1.0-L vessel at-80 C and let equilibrium be established. The total pressure in the system at equilabrium will be: A) 0.79 atm (B1 2.3 atm C1 0.23 atmDJ 1.3 atm [EJ none of these 21. The reaction has Kp 45.9 at 763 K. A particular equilibrium mixture at that temperature contains

Explanation / Answer

Answer

C) 5.4×10-13

Explanation

H2(g) + O2(g) <- - - - - - > H2O2(g) Kp = 2.3×106 - - - - eq(1)

2H2(g) + O2(g) <- - - - - - > 2H2O(g) Kp = 1.8×1037 - - - - - eq(2)

multiply eq1 by 2

2H2(g) + 2O2(g) < - - - > 2H2O2(g) Kp = 5.29×1012- - - - eq(3)

reverse eq(2)

2H2O <- - - - - - > 2H2(g) + O2(g) Kp = 5.6×10-38 ---- eq(4)

add eq(3) and eq(4)

2H2(g) + 2O2(g) + 2H2O(g) <- - - > 2H2O2(g) + 2H2(g) + O2(g) Kp =2.96×10-25- - - eq(5)

Divide eqn5 divide by 2

H2O(g) + 1/2O2(g) < - - - > 2H2O2(g) Kp = 5.44×10-13

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