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Chemistry, 29.11.2019 17:31 Bbmonstr8279

The kp for the reaction below is 1.49 × 108 at 100.0 °c:

co (g) + cl2 (g) cocl2 (g)

in an equilibrium mixture of the three gases, pco = pcl2 = 1.00 × 10-4 atm. the partial pressure of the product, phosgene (cocl2), is atm.

Answers

ansver
Answer from: kieante01
The equilibrium constant in this case is [CO2]/[Cl2][CO]. Given the Kp, we can get directly the partial pressure of phosgene which is 1.49x10^8*1x10^-4*1x10^-4. This is equal to 1.49 atm. 
ansver
Answer from: Quest

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

ansver
Answer from: Quest

the temperature of the gas, in k, not celsius.

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The kp for the reaction below is 1.49 × 108 at 100.0 °c:

co (g) + cl2 (g) cocl2 (g)