What change would shift the equilibrium system to the left? A(g) + B(s) + Energy ⇌ 3C(g) (3 points)
Adding more of gas C to the system
Heating the system
Increasing the volume
Removing some of gas C from the system

A.) Adding more of gas C to the system

Explanation:

A.) is correct because the reaction wants to keep the same amount of reactants and products that sustain equilibrium. If you are increasing the amount of gas C, this will push the reaction to make more of the reactants to restore the balance.

B.) is incorrect because heat is a form of energy. As previously stated, when you add more material to the reactants side, the equilibrium shifts to the product side to restore equilibrium.

C.) is incorrect because when the volume is increased, the equilibrium will shift to the side with the greater amount of moles of gas. There is 1 mole of gas on the reactant side (from A (g)) and 3 moles of gas on the products side (from 3 C (g)). Therefore, the reaction will shift to the product side.

D.) is incorrect because if the material on the product side is decreasing, the equilibrium will want to make up that difference by shifting to the product side.

Remember,

reactants ---> products

explanation: no but i do not usually comment

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