Question

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tion 18 O out of 4 points At the start of a reaction, there are 1.05 moles of NO, 0.955 moles of Bry, and 1.15 moles of NOBr in a 5.5 L reaction vessel at 727 "C. From these initial molar concentrations, decide whether the system is at equilibrium. If not, predict which direction the net reaction will proceed. (Hint: Reaction Quotient) 2NO(g) + Br2(g) + 2NOBr() Ke 0.013 Question 19 O out of 4 points At a temperature of 700 °C the equilibrium constant value (c) for the following reaction equal to 1,08 * 107. An equilibrium mixture was found to have the following concentrations of H2 and H2S: [H21 -2.25 M and [H251 - 3.15 M. What is the molar concentration of S218) at equilibrium, 152] - 2H2(g) +S2(g) =2H2S(g) Kc = 1.08 107 Question 20 out of 4 points At a temperature of 800 °C the equilibrium constant value (Kp) for the following reaction equals to 2.5 * 109. An equilibrium mixture was found to have the following partial pressure for O2 and SOz: PO2) - 105 atm and P4SO3) - 211 atm. What is the partial pressure of SO2(g) at equilibrium, PISO2) 2SO2(e) + O2(g) 2503(g) Kp -2.5 * 109 Question 21 O out of 4 points Chemical analysis of a 6.5 L container found that at equilibrium there are 0.344 moles of O3(e) and 0.688 moles of O2(8). Using this information, what is the equilibrium constant (Kc) for the following reaction at 300 °C? [203(8) 302(8) ke-?)

Answer 1

18 Given reaction is Kc = 0.013 2NO(g) +B7,49) Q NO BOLS) moles of NOS 1.05 mol moles of Br₂ = 0.955 mol moles of NOBT = 115 mol volume of container 5.5L concertation moles Volune we know that reaction quotient 2 Also by law of mass ochon C'NOBY [no] Br] 1.15 ml or 7.05 mol up on calculations kg 6c 908 6.91 Katke Peaction equilibrium بل nor Reaction direction will proceed is backward curtain equilibrium. Given reaction is 2 H3 (9) + Solg) 2119569) K=1.08 XI0 Given equilibrium concentration of [H,3 = 2,25M [125) = 3.15 M [52] = ? By law of mass action Scanned with CamScanner

Kes CH25) [H2] ?CS, 1.08x10? (3.15)? (2.25)x ($2] upon calculations, we get [S2] = 1.81 X10-7 M 00 Ginen reaction is 250, (9) + O2(g) 2503 (9) Kp = 2.5 X109 Ginen equilibrium partial pressures of PCO2) = 105 am P(S) = 2liam P(50) ? By law of man achen of equilibrium Kp = P (50₂) Pt50)+10 or 2.5X1095 T2112 crosjex PCSO)? upon calculations PC50,) = 1.696X10 170 x 10-7 P (50,)= 4.12X10 am S Scanned with CamScanner

Given, moles of o 03-0.344 mal moles of o₂ = 0.688 mol volume of containese = 6.5L Reaction is 20,69) 230,(9) By law of mass acbon ke = [0₂] 0.688 [ ₃ ]² 0.344 at equilibrium 69.688 3 upon calculations Kel 0 2 3 CS Scanned with CamScanner
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