Open application with bundle identifier

You can use private API to do that

Class LSApplicationWorkspace_class = objc_getClass("LSApplicationWorkspace");
NSObject * workspace = [LSApplicationWorkspace_class performSelector:@selector(defaultWorkspace)];
BOOL isopen = [workspace performSelector:@selector(openApplicationWithBundleID:) withObject:@"com.apple.mobilesafari"];

You can use the openUrl call, but in order to succeed you must add some values to your project's xy-Info.plist file.

Once you've done that you can then call:

[[UIApplication sharedApplication] openUrl:[NSURL urlWithString:@"xingipad://"]];

I don't think that's possible.