Answer
b, d and e will cause the equilibrium to shift to the right.
Work Step by Step
a) The number of moles of gas is lower for the reactants (1). Therefore, decreasing the volume will shift the equilibrium to the left.
b) The number of moles of gas is greater for the products (2). Therefore, increasing the volume will shift the equilibrium to the right.
c) C(graphite) is a solid, so it doesn't affect the equilibrium.
d) $CO_2$ is a reactant, so the addition of $CO_2$ will shift the equilibrium to the right.
e) $CO$ is a product, so the removal of $CO$ will shift the equilibrium to the right.
